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USEMO-2019-1
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https://web.evanchen.cc/exams/report-usemo-2019.pdf
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Let \(ABCD\) be a cyclic quadrilateral. A circle centered at \(O\) passes through \(B\) and \(D\) and meets lines \(BA\) and \(BC\) again at points \(E\) and \(F\) (distinct from \(A, B, C\)). Let \(H\) denote the orthocenter of triangle \(DEF\). Prove that if lines \(AC, DO, EF\) are concurrent, then triangles \(ABC\) and \(EHF\) are similar.
|
Define \(G\) as the intersection of \(\overline{AC}\) and \(\overline{EF}\).
extbf{Claim} â Quadrilateral \(DCGF\) is cyclic.
extit{First proof.} Because
\[ ngle DCG = ngle DCA = ngle DBA = ngle DBE = ngle DFE = ngle DFG. \]
extit{Second proof.} Follows since \(D\) is Miquel point of \(GABF\).
extbf{Claim} â If \(G\) lies on line \(\overline{DO}\), we have \(\overline{AC} \perp \overline{BD}\).
extit{Proof.} We have
\[ ngle BDG = ngle BDO = 90^\circ - ngle DEB = 90^\circ - ngle DFB
= 90^\circ - ngle DFC = 90^\circ - ngle DGC. \]
\(\square\)
To finish,
\[ ngle HEF = 90^\circ - ngle EFD = 90^\circ - ngle EBD = 90^\circ - ngle ABD = ngle CAB. \]
Similarly \(ngle HFE = ngle ACB\) and the proof is done.
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Most solutions are worth 0 or 7. The following partial items are available but extit{not additive}:
- 5 points for proving if \(AC, DO, EF\) are concurrent implies \(AC \perp BD\) but not finishing.
- 4 points for solutions for which spiral similarity justification is entirely absent, but which would be complete if these details were supplied correctly.
- 1 point for proving that \(AC \perp BD\) is equivalent to the problem.
There is no deduction for configuration issues (Such as not using directed angles) or small typos in angle chasing.
No points awarded for noting \(ngle ABC = ngle EHF\), or proving/noting that \(D\) is the Miquel point of \(AEFC\) but not making further progress on the problem.
Computational approaches which are not completed are judged by any geometric content and do not earn other marks.
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USEMO-2019-2
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https://web.evanchen.cc/exams/report-usemo-2019.pdf
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Let \( \mathbb{Z}[x] \) denote the set of single-variable polynomials in \( x \) with integer coefficients. Find all functions \( heta : \mathbb{Z}[x] o \mathbb{Z}[x] \) (i.e. functions taking polynomials to polynomials) such that
- for any polynomials \( p, q \in \mathbb{Z}[x] \), \( heta(p + q) = heta(p) + heta(q) \);
- for any polynomial \( p \in \mathbb{Z}[x] \), \( p \) has an integer root if and only if \( heta(p) \) does.
|
The answer is that
\[ heta(x) = r(x) \cdot p(\pm x + c) \]
for any choice of \( c \in \mathbb{Z} \), \( r(x) \) without an integer root, with the choice of sign fixed. For the converse direction we present two approaches.
**First solution.**
Let \( r(x) = heta(1) \), which has no integer root since the constant 1 has no roots at all.
**Part 1.** We fix a positive integer \( n \) and start by determining \( heta(x^n) \). Let \( f(x) = heta(x^n) \). We look at
\[ heta(ax^n + b) = a \cdot f(x) + b \cdot r(x). \]
Let \( g(x) = f(x)/r(x) \), a quotient of two polynomials whose denominator never vanishes. By using the problem condition in both directions, varying \( x \in \mathbb{Z} \) and \( -b/a \in \mathbb{Q} \), we find that
\[ rac{f(x)}{r(x)} ext{ takes on exactly the values } \ldots, (-2)^n, (-1)^n, 0^n, 1^n, 2^n, 3^n, \ldots ext{ for } x \in \mathbb{Z}. \]
So let \( g(x) = f(x)/r(x) \) now.
**Claim (Rational functions canât be integer-valued forever)** â Since \( g \) maps integers to integers, it must actually be a polynomial with rational coefficients.
**Proof.** We will only need the condition that \( g \) maps integers to integers. If not, then by the division algorithm, we have \( g(x) = d(x) + rac{f_1(x)}{f_2(x)} \) for some polynomials \( d(x), f_1(x), f_2(x) \in \mathbb{Q}[x] \) with \( \deg f_2 > \deg f_1 \ge 0 \). There exists an integer \( D \) such that \( D \cdot d(x) \in \mathbb{Z}[x] \) (say the lcm of the denominators of the coefficients of \( g \)). But for large enough integers \( x \) the value of \( rac{f_1(x)}{f_2(x)} \) is a nonzero and has absolute value less than \( rac{1}{D} \). This is a contradiction. \( \square \)
**Remark.** You canât drop the condition that \( g \) has rational (rather than integer) coefficients in the proof above; consider \( g(x) = rac{1}{2}x(x+1) \) for example.
Let \( C \) be an integer divisible by every denominator in the coefficients of \( g \). Then apparently
\[ h(x) = C^n \cdot g(x) \]
is a polynomial which only takes on \( n \)th powers as \( x \in \mathbb{Z} \).
**Claim (Polya and Szego)** â Since \( h \) is a polynomial with integer coefficients whose only values are \( n \)th powers, it must itself be the \( n \)th power of a polynomial.
**Proof.** This is a classical folklore problem, but we prove it for completeness. Decompose \( h \) into irreducible factors as
\[ h(x) = c \cdot f_0(x)^{e_0} \cdot f_1(x)^{e_1} \cdot f_2(x)^{e_2} \cdot f_3(x)^{e_3} \cdots f_m(x)^{e_m} \]
where the \( f_i \) are nonconstant and \( c \) is an integer, and \( e_i > 0 \) for all \( i > 0 \). We also assume \( m > 0 \).
We use the following facts:
- In general, if \( A(x), B(x) \in \mathbb{Z}[x] \) are coprime, then \( \gcd(A, B) \) is bounded by some constant \( C_{A,B} \). This follows by Bezout lemma.
- If \( A(x) \in \mathbb{Z}[x] \) is a nonconstant polynomial, then there are infinitely many primes dividing some element in the range of \( A \). This is called Schurâs theorem.
- Let \( A(x) \in \mathbb{Z}[x] \) be an irreducible polynomial, and let \( A'(x) \) be its derivative. Then if \( p \) is prime and \( p > C_{A,A'} \), and \( p \) has root in \( \mathbb{F}_p \), then there exists \( x \) with \(
u_p(A(x)) = 1 \). This follows by Hensel lemma.
Now for the main proof. By the above facts and the Chinese remainder theorem (together with Dirichlet theorem), we can select enormous primes \( p_1 < p_2 < \cdots < p_m < q \) (exceeding \( c, e, \max e_i, \max C_{f_i,x}, \max C_{f_i,f_j} \) for all \( i \) and \( j \)) and a single integer \( N \) satisfying the following constraints:
- \(
u_{p_i}(f_i(N)) = 1 \) for all \( i = 1, \ldots, m \), by requiring \( N \equiv t_i \pmod{p_i^2} \) for suitable constant \( t_i \) not divisible by \( p_i \) (because of Hensel lemma);
- \( p_i
mid f_j(N) \) whenever \( i
e j \); this follows by the fact that \( p_i > C_{f_i,f_j} \).
Now look at the value of \( f(N) \). It has
\[
u_{p_1}(f(N)) = e_1 \
u_{p_2}(f(N)) = e_2 \
dots \
u_{p_m}(f(N)) = e_m. \]
Now \( f(N) \) is an \( n \)th power so \( n \) divides all of \( e_1, \ldots, e_m \). Finally \( c \) must be an \( n \)th power too. \( \square \)
So \( h(x) \) is an \( n \)th power; thus so is \( g(x) \). Letâs write \( g(x) = p(x)^n \) then; so we find that the range of \( p(x) \) contains either \( k \) or \( -k \), for every integer \( k \). For density reasons, this forces \( p \) to be linear, and actually of the form \( p(x) = \pm x + c \) for some constant \( c \).
**Part 2.** We have now shown \( heta(x^n) = (\pm x + c)^n r(x) \), for every \( n \), for some sign and choice of \( c \) depending possibly on \( n \). It remains to show that the choices of signs and constants are compatible across the different values of \( n \). So letâs verify this.
By applying a suitable transformation on \( x \) letâs assume \( heta(x) = x \) for simplicity. Then look at \( heta(x^n + ax) = (\pm x + c)^n + ax \) for choices of integers \( a \). This is apparently supposed to have a root for each choice of \( a \), but if \( c
e 0 \), this means \( rac{1}{x}(\pm x + c)^n \) can take any integer value, which is obviously not true for density reasons. This means \( c = 0 \), so it shows \( heta(x^n) = \pm x^n \) for any integer \( n \).
Finally, by considering \( heta(x^n + x - 2) = \pm x^n - x + 2 \), we see the sign must be \( + \) for the RHS to have an integer root. This finishes the proof.
**Second solution, outline (by contestants).**
**Claim (Odd-degree polynomials are determined by their range)** â Let \( P(x) \in \mathbb{Z}[x] \) be an odd-degree polynomial. Let \( Q(x) \) be another polynomial with the same range as \( P \) over \( \mathbb{Z} \). Then \( P(x) = Q(\pm x + c) \) for some \( \pm 1 \) and \( c \).
**Proof.** First, \( Q \) also has odd degree since it must be unbounded in both directions. By negating if needed, assume \( Q \) has positive leading coefficient.
Take a sufficiently large integer \( n_0 \) such that \( P(x) \) and \( Q(x) \) are both strictly increasing for \( x \ge n_0 \), and moreover \( P(n_0) > \max_{x < n_0} P(x), Q(n_0) > \max_{x < n_0} P(x) \). Then take an even larger integer \( n_1 > n_0 \) such that \( \min(P(n_1), Q(n_1)) > \max(P(n_0), Q(n_0)) \). Choose \( n_2 > n_0 \) such that \( P(n_1) = Q(n_2) \). We find that this implies
\[ P(n_1) = Q(n_2) \
P(n_1 + 1) = Q(n_2 + 1) \
P(n_1 + 2) = Q(n_2 + 2) \
P(n_1 + 3) = Q(n_2 + 3) \]
and so on. So \( P \) is a shift of \( Q \) as needed. \( \square \)
This is enough to force \( heta(x^n) = (\pm x + c)^n r(x) \) when \( n \) is odd. When \( n \) is even, for each integer \( k \) one can consider
\[ heta(kx^{n+3} + x^n) = k heta(x^{n+3}) + heta(x^n) \]
and use the claim on \( heta(x^{n+3}) \) and \( heta(kx^{n+3} + x^n) \) to pin down \( heta(x^n) \).
**Third solution (from author).**
**Lemma**
Given two polynomials \( P, Q \in \mathbb{Z}[x] \), if \( P + nQ \) has an integer root for all \( n \), then either \( P \) and \( Q \) share an integer root or \( P(x) = \left( rac{x + m}{k}
ight) Q(x) \) for some integers \( m, k \) with \( k
e 0 \).
**Proof.** Let \( d = \gcd(P(0), Q(0)) \) so \( P(0) = dr \) and \( Q(0) = ds \). Now, for an integer root \( k_n \) of \( P + nQ \),
\[ k_n | P(0) + nQ(0) = dr + nds = d(r + ns). \]
Let \( p \) be a prime \( \equiv r \mod s \), of which there are infinitely many by Dirichletâs theorem. Now, for \( n = rac{p - r}{s} \), we have
\[ k_n | dp. \]
As the divisors of \( dp \) are exactly those of \( d \) times 1 or \( p \), there exists a (not necessarily positive) divisor \( j \) of \( d \) and a \( t \in \{1, p\} \) so that \( k_n = dt \) for infinitely many \( n \). In the first case, we have that \( P(j) + nQ(j) = 0 \) for infinitely many \( n \) and some fixed \( j \), which implies that \( j \) is a root of both \( P \) and \( Q \). In the second case, we have, noting \( p = r + ns \), that
\[ P(j(r + ns)) + nQ(j(r + ns)) = 0. \]
As this holds for infinitely many \( n \), we may rewrite it as a polynomial equation
\[ P(x) = (ax + b)Q(x) \]
for some rational \( a, b \). Now, we know that \( (ax + b + n)Q(x) \) has a rational root for all \( n \in \mathbb{Z} \). If \( Q \) has an integer root then \( P \) does as well and we are in our first case; otherwise, \( rac{n b}{a} \in \mathbb{Z} \) for all \( n \in \mathbb{Z} \). This implies that \( 1/a \in \mathbb{Z} \), let it be \( k \). Then \( b/a \in \mathbb{Z} \); let it be \( m \).
Now, let \( P_n(x) = f(x^n) \). We claim that \( P_1(x) = (\pm x + t)P_0(x) \) for some \( t \in \mathbb{Z} \). Indeed, \( P_1 + nP_0 \) has an integer root for all \( n \), so either \( P_1 \) and \( P_0 \) share an integer root or \( P_1(x) = \left( rac{x + m}{k}
ight) P_0(x) \) for some \( m, k \in \mathbb{Z} \). They clearly cannot share a root, since \( P_0(x) \) cannot have any integer roots. Now,
\[ kP_1(x) + P_0(x) = (x + m + k)P_0(x) \]
has an integer root, so \( kx + 1 \) must as well, and thus \( k = \pm 1 \), as desired. Now, we see that
\[ heta\left(a(x^n - c^n) + b(x - c)
ight) = a\left(P_n(x) - c^n P_0(x)
ight) + b\left(P_1(x) - cP_0(x)
ight) \]
has an integer root for any \( c, a, b \). Let \( Q = P_n - c^n P_0 \) and \( R = P_1 - cP_0 \). Since \( aQ + bR \) has an integer root for all \( a, b \in \mathbb{Z} \), we can apply our lemma on both the pair \( (Q, R) \) and \( (R, Q) \); if they do not share an integer root, then \( Q \) must be a linear polynomial times \( R \) and \( R \) must be a linear times \( Q \), a contradiction unless they are both 0 (in which case they share any integer root). So, \( Q \) and \( R \) share an integer root. We have
\[ R(x) = P_1(x) - cP_0(x) = (\pm x + t - c)P_0(x), \]
and \( P_0 \) has no integer root as 1 has no integer root, so we have that \( \pm(c - t) \) is the only integer root of \( R \) and is thus also a root of \( Q \); in particular
\[ P_n(\pm(c - t)) = c^n P_0(\pm(c - t)) \]
for all \( c \in \mathbb{Z} \). This is a polynomial equation that holds for infinitely many \( c \) so we must have that
\[ P_n(\pm(x - t)) = x^n P_0(\pm(x - t)) \implies P_n(x) = (\pm x + t)^n P_0(x). \]
Thus, if \( Q(x) = \sum_{i=0}^d a_i x^i \),
\[ heta(Q(x)) = heta\left(\sum_{i=0}^d a_i x^i
ight) = \sum_{i=0}^d a_i (\pm x + t)^i P_0(x) = P_0(x) Q(\pm x + t), \]
finishing the proof.
|
For solutions which are not complete, the following items are available but *not additive*:
- 0 points for the correct answer.
- 1 point for proving \( heta(1) \) divides \( heta(P) \) over \( \mathbb{Q}[x] \)
- 1 point for showing \( heta(x), heta(x^2), \ldots \) all have a common integer root, or that every pair does.
- 1 point for showing \( heta(x)/ heta(1) \) is a linear polynomial with rational coefficients.
- 2 points are awarded for a solution that starts to make progress on \( heta(x^n) \) for \( n \ge 2 \), by proving some main lemma or claim. Showing that \( heta(x) \) is a linear multiple of \( heta(1) \) does not earn this point.
(A common wrong approach is to claim that the rational function \( rac{ heta(x^n)}{ heta(x^{n-1})} \) takes on every integer value; this does not work since \( heta(x^n) \) and \( heta(x^{n-1}) \) could have a common root for \( n \ge 2 \).)
For solutions which are complete with errors, the following deductions apply, and all deductions are additive:
- â1 point for an incorrect answer. This may include forgetting the \( \pm 1 \), for example.
- â1 point for a minor error. This most commonly applies to students who took the quotient of two integer polynomials and assumed the coefficients were integers when in fact they could be rational numbers, but the solution can be easily patched once this is pointed out.
- â2 points for a more significant error that is easily fixable.
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USEMO-2019-3
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https://web.evanchen.cc/exams/report-usemo-2019.pdf
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Consider an infinite grid \( \mathcal{G} \) of unit square cells. A \emph{chessboard polygon} is a simple polygon (i.e. not self-intersecting) whose sides lie along the gridlines of \( \mathcal{G} \).
Nikolai chooses a chessboard polygon \( F \) and challenges you to paint some cells of \( \mathcal{G} \) green, such that any chessboard polygon congruent to \( F \) has at least 1 green cell but at most 2020 green cells. Can Nikolai choose \( F \) to make your job impossible?
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The answer is YES, the task can be made impossible.
The solution is split into three parts. First, we describe a âpolygon with holesâ \( F \). In the second part we prove that this \( F \) works. Finally, we show how to take care of the holes to obtain a true polygon.
**Part 1. Construction.** Choose large integers \( m \geq 5, n \geq 1 \) with \( m \) odd. We will let
\[ s = m^n \]
throughout the solution.
Let \( F_0 \) be a square of side length \( s \). Starting from \( F_0 \), iterate the following procedure. Divide \( F_i \) into squares of side \( s/m^i \) and poke a square hole of side \( 3s/m^{i+1} \) (a \emph{phase-\( i \)} hole) in the center of each such square to obtain \( F_{i+1} \). Finally, let \( F = F_n \).
The output is shown below for \( m = 11 \) and \( n = 3 \). The claim is that for a suitable choice of \( (m,n) \) this will serve as the desired example.
**Part 2. Proof of this example.** Suppose we have a green coloring as described. For every green cell, the \emph{standard copy} of \( F \) is a copy of \( F \) centered at the green cell.
**Claim.** The standard copies of \( F \) completely cover the plane.
**Proof.** This is equivalent to every copy of \( F \) having at least one green cell, owing to symmetry of \( F \). \( \square \)
**Claim.** In any square \( C \) with side length \( 5s \), there are at least \( n+1 \) standard copies of \( F \).
**Proof.** Let \( C_0 \) be a square of side \( 3s \) concentric with \( C \). Starting from \( C_0 \), iterate the following procedure for \( 1 \leq i \leq n+1 \):
- Let \( S_{i-1} \) be one of the standard copies of \( F \) that cover the center of \( C_{i-1} \),
- If \( i
e n+1 \), let \( C_i \) be a phase-\( i \) hole in \( S_{i-1} \) that lies in the interior of \( C_{i-1} \).
Since \( C_0 \) and the holes \( C_1, C_2, \ldots, C_n \) are nested, the standard copies \( S_0, S_1, S_2, \ldots, S_n \) of \( F \) are distinct. It follows that \( C \) contains at least \( n+1 \) standard copies of \( F \). \( \square \)
However, the area of \( F \) is
\[ s^2 \left(1 - rac{9}{m^2}
ight)^n. \]
Therefore, at least one cell \( c \) within \( C \) is covered by at least
\[ k = rac{n+1}{25} \left(1 - rac{9}{m^2}
ight)^n \]
standard copies of \( F \). The copy of \( F \) centered at \( c \) then contains at least \( k \) green cells.
When \( n = 25 \cdot 2020 \) and \( m \) is sufficiently large, however, we get \( k > 2020 \).
**Part 3. Handling the holes.** We are left to show how to repair the above construction so that \( F \) becomes a true polygon.
Let \( D \) be any sufficiently large positive integer. Consider a homothetic copy \( F_D \) of \( F \) scaled by a factor of \( D \). Cut several canals of unit width into \( F_D \) so that \( F_D \) continues to be connected and every hole in \( F_D \) is joined by a canal to the boundary of \( F_D \). (Canals do not need to be straight; they may go around holes.) When \( F_D \) is repaired in this way, it becomes a true polygon \( F_D' \).
Since the total area of all canals is proportional to \( D \) and the total area of \( F_D \) is proportional to \( D^2 \), when \( D \) becomes arbitrarily large the ratio of the area of \( F_D' \) to the area of \( F_D \) becomes arbitrarily close to one. Therefore, for all sufficiently large \( D \) our proof that \( F \) is in fact a counterexample goes through for \( F_D' \) as well, with straightforward adjustments.
The solution is complete.
|
- 0 points for the correct answer
- 6 points for a solution that uses a set of grid cells which do not form a polygon, but is otherwise correct (this includes sets which are not connected)
- 7 points for a correct solution
Any partial credit is done on case-by-case basis.
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USEMO-2019-4
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https://web.evanchen.cc/exams/report-usemo-2019.pdf
|
Prove that for any prime \( p \), there exists a positive integer \( n \) such that
\[
1^n + 2^{n-1} + 3^{n-2} + \cdots + n^1 \equiv 2020 \pmod{p}.
\]
|
The idea is to pick \( n = c \cdot p \cdot (p - 1) \) for suitable integer \( c \). In what follows, everything is written modulo \( p \).
**Claim** â When \( n = c \cdot p \cdot (p - 1) \), the left-hand side is equal to
\[
c \cdot \sum_{a=0}^{p-2} \sum_{b=1}^{p-1} b^a = c \cdot \left[1^0 + 2^0 + \cdots + (p-1)^0 \
+ 1^1 + 2^1 + \cdots + (p-1)^1 \
+ 1^2 + 2^2 + \cdots + (p-1)^2 \
+ \cdots \
+ 1^{p-2} + 2^{p-2} + \cdots + (p-1)^{p-2}
ight].
\]
*Proof.* In the original sum, we discard all the terms divisible by \( p \), reduce all the bases modulo \( p \), and reduce all the exponents modulo \( p - 1 \) (by Fermatâs little theorem). Then each block of \( p(p - 1) \) terms equals
\[
1^0 + 2^{p-2} + 3^{p-3} + \cdots + (p - 1)^1 \
+ 1^2 + 2^{p-3} + 3^{p-4} + \cdots + (p - 1)^0 \
+ 1^3 + 2^{p-4} + 3^{p-5} + \cdots + (p - 1)^{p-2} \
+ \cdots \
+ 1^1 + 2^0 + 3^{p-2} + \cdots + (p - 1)^2
\]
which rearranges to the desired sum. \( \square \)
**Claim** â We have
\[
\sum_{a=0}^{p-2} \sum_{b=1}^{p-1} b^a \equiv -1 \pmod{p}.
\]
*First proof.* By the geometric series formula
\[
\sum_{a=0}^{p-2} b^a = rac{b^{p-1} - 1}{b - 1} = 0 \quad orall b = 2, 3, \ldots, p - 1.
\]
The terms with \( b = 1 \) contribute \( 1^0 + 1^1 + \cdots + 1^{p-2} = p - 1 \) and done. \( \square \)
*Second proof.* In fact, itâs a classical lemma (proved in the same way, using primitive roots) that
\[
\sum_{b=1}^{p-1} b^a \equiv egin{cases}-1 & p - 1 \mid a \ 0 & p - 1
mid a \end{cases} \pmod{p}
\]
so this is immediate. \( \square \)
Thus we simply need to select \( c \equiv -2020 \pmod{p} \) and win (and \( c > 0 \)).
|
For solutions which are not complete, the following items apply but are not additive:
- 0 points for no progress.
- 0 points for considering \( n = cp(p - 1) \) and nothing else.
- 0 points for partial steps in computing this sum, e.g. using primitive roots or the geometric series but not tying it together.
- 1 point for showing the sum when \( n = cp(p - 1) \) is \( c \) times the sum when \( n = p(p - 1) \), but not evaluating the latter sum.
- 1 point for evaluating the sum when \( n = p(p - 1) \) but not finishing the problem.
For essentially complete solutions, the following deductions could apply and are additive:
- \(-1\) point: the student shows how to solve the problem for any negative number in place of 2020, but doesnât realize you can wrap around.
- \(-1\) point: the student calculates the sum at \( n = p(p - 1) \) and from this assumes that the sum of any \( n \) consecutive terms is \( -1 \mod p \).
- \(-1\) point: the student states with no proof that the sum of \( x^k \) as \( x \) varies across a residue system mod \( p \) is \( -1 \) if \( p - 1 \mid k \) and 0 otherwise. (Mentioning primitive roots is OK.)
- \(-1\) point: calculation error leading to wrong final answer
Do not deduct for the following errors:
- The student forgets things like \( 0^{p-1} = 0 \) instead of 1, as long as the errors do not change the final result.
- The student uses negative exponents on multiples of \( p \), as long as the solution would be correct if the negative powers were replaced with âcorrespondingâ positive powers
|
USEMO-2019-5
|
https://web.evanchen.cc/exams/report-usemo-2019.pdf
|
Let \( P \) be a regular polygon, and let \( \mathcal{V} \) be the set of its vertices. Each point in \( \mathcal{V} \) is colored red, white, or blue. A subset of \( \mathcal{V} \) is \emph{patriotic} if it contains an equal number of points of each color, and a side of \( P \) is \emph{dazzling} if its endpoints are of different colors.
Suppose that \( \mathcal{V} \) is patriotic and the number of dazzling edges of \( P \) is even. Prove that there exists a line, not passing through any point of \( \mathcal{V} \), dividing \( \mathcal{V} \) into two nonempty patriotic subsets.
|
We prove the contrapositive: if there is no way to split \( \mathcal{V} \) into two patriotic sets, then the number of dazzling edges is odd.
Let \( \zeta = - rac{1}{2} + rac{\sqrt{3}}{2}i \) be a root of unity. Read the \( n \) vertices of the polygon in order starting from any point. In the complex plane, start from the origin and, corresponding to red, white, or blue, move by \( 1 \), \( \zeta \), or \( \zeta^2 \), respectively, to get a path. The diagram below shows an example (where black stands in for white, for legibility reasons).
Note that:
- The path we get is actually a closed loop, since \( \mathcal{V} \) was assumed to be patriotic.
- Because there is no nontrivial patriotic subset, this closed loop does not intersect itself, so it corresponds to some polygon \( Q \).
We have to show the number \( m \) of vertices of \( Q \) (corresponding to dazzling edges) is odd.
Let \( x \) and \( y \) denote the number of \( 60^\circ \) and \( 300^\circ \) angles, so
\[
60x + 300y = 180(x + y - 2).
\]
This gives \( x - y = 3 \) so \( x + y \) is odd.
|
⢠0 points for an incorrect solution.
⢠0 points for a 1-dimensional tracker argument, to prove e.g. the problem with two colors instead of three.
⢠1 point for making a broken-line polygon (with any set of angles) and not finishing. This includes both:
â Making a polygon using 1, \( \omega \), \( \omega^2 \) vectors and not attempting the angle-sum argument.
â Making a polygon with a different set of vectors summing to 0, such that finishing with the angle-sum argument is hard.
⢠6 points for a correct solution with a minor error.
The most common form of this will likely be the official solution with a different set of vectors, with a mistake in the resulting algebra. The mistake must be easily fixable.
⢠7 points for a correct solution.
A number of solutions make a polygon with 1, \( \omega \), \( \omega^2 \) vectors and then try to âsmooth awayâ 300 degree angles. For this to be graded 7â, the smoothing argument must be very explicit. It should be able to handle extremely concave shapes, like the one below. Otherwise, just award the 1 point for considering the broken-line polygon.
|
USEMO-2019-6
|
https://web.evanchen.cc/exams/report-usemo-2019.pdf
|
Let \( ABC \) be an acute scalene triangle with circumcenter \( O \) and altitudes \( AD, BE, CF \). Let \( X, Y, Z \) be the midpoints of \( AD, BE, CF \). Lines \( AD \) and \( YZ \) intersect at \( P \), lines \( BE \) and \( ZX \) intersect at \( Q \), and lines \( CF \) and \( XY \) intersect at \( R \).
Suppose that lines \( YZ \) and \( BC \) intersect at \( A' \), and lines \( QR \) and \( EF \) intersect at \( D' \). Prove that the perpendiculars from \( A, B, C, O \) to the lines \( QR, RP, PQ, A'D' \), respectively, are concurrent.
|
Solution 1
(Radical axis approach)
The main idea is to show that \( (DEF) \) and \( (XYZ) \) has radical axis \( A'D' \).
Let \( H \) be the orthocenter of \( riangle ABC \). We'll let \( (AH), (BH), (CH) \) denote the circles with diameters \( AH, BH, CH \).
extbf{Claim.} Points \( H, D, Y, Z \) are cyclic.
extit{Proof.} Let \( M \) be the midpoint of \( BC \). We claim they lie on a circle with \( HM \).
Clearly \( ngle HDM = 90^\circ \). The segment \( YM \) is the \( B \)-midline of \( riangle BEC \), so \( YM \parallel EC \perp HY \): thus \( ngle HYM = 90^\circ \). Similarly \( ngle HZM = 90^\circ \). \( \square \)
extbf{Claim.} The point \( P \) is the radical center of \( (HB), (HC), (XYZ), (HYZD) \). Also, \( QR \) is the radical axis of \( (HA) \) and \( (XYZ) \).
extit{Proof.} First part since \( PH \cdot PD = PY \cdot PZ \); second part by symmetric claims. \( \square \)
We are now ready for the key claim.
extbf{Claim (Key claim).} The points \( A' \) and \( D' \) lie on the radical axis of \( (DEF) \) and \( (XYZ) \).
extit{Proof.} The radical center of \( (DEF), (XYZ), (HYZD) \) is \( A' = YZ \cap BC \), and the radical center of \( (DEF), (XYZ), (HA) \) is \( D' = EF \cap QR \), so we're done. \( \square \)
Let \( S \) be the center of \( (XYZ) \) and \( T \) the reflection of \( H \) over \( S \). Let \( N \) denote the nine-point center.
extbf{Claim (Concurrence).} The point \( T \) is the concurrency point in the problem.
extit{Proof.} The line through the centers of \( (HA) \) and \( (XYZ) \) is perpendicular to the radical axis \( QR \). Now, a homothety with center \( H \) and scale 2 sends these centers to \( A \) and \( T \), so \( AT \perp QR \). Similarly, \( BT \perp RP \) and \( CT \perp PQ \).
Similarly from \( NS \perp A'D' \), a dilation at \( H \) by a factor of 2 shows \( OT \perp A'D' \), as desired. \( \square \)
extbf{Remark.} (Author comments on problem creation). The main goal was to create a problem to showcase the midpoints of the altitudes: while they arise due to the midpoint of altitude lemma (Lemma 4.14 in EGMO), I have rarely seen them studied in their own right. This problem strives to be a synthesis of properties relating to the midpoints of altitudes.
extbf{Remark.} An original, more long-winded version of the problem asks to show that if \( B', C', E', F' \) are defined similarly, then all six points are collinear and perpendicular to \( OT \). The second approach below proves this.
Solution 2
(Orthology approach)
Define \( B', C', E', F' \) in an analogous fashion.
extbf{Claim.} Points \( A', B', C', D', E', F' \) are collinear.
extit{Proof.} Three applications of Desargue:
- \( ABC \) and \( XYZ \) are perspective at \( H \) so \( A', B', C' \) are collinear.
- \( DEF \) and \( PQR \) are perspective at \( H \) so \( D', E', F' \) are collinear.
- \( C'FR \) and \( B'EQ \) are perspective through \( A \)-altitude so \( B'C', EF, QR \) are concurrent (at \( D' \)). \( \square \)
extbf{Claim.} The perpendiculars from \( A, B, C \) to \( QR, RP, PQ \) are concurrent.
extit{Proof.} This follows from the fact that \( riangle ABC \) and \( riangle PQR \) are orthologic with one orthology center at \( O \). \( \square \)
extbf{Claim.} The perpendiculars from \( A, O, C \) to \( QR, D'F', PQ \) are concurrent.
extit{Proof.} This follows from the fact that \( riangle D'F'Q \) and \( riangle AOC \) are orthologic with one orthology center at \( E \) (note that \( AO \perp EF \)). \( \square \)
extbf{Remark.} This solution does not even use the fact that \( X, Y, Z \) were the midpoints of the altitudes!
|
As this problem is difficult, there are not many correct solutions, so we may manually flag any unusual cases to review as a group. Hence, this rubric is very sparse on details, outlining only the common 0⺠or 7⻠cases.
The following partial items are not additive:
- 0 points for proving just the concurrence of the first three perpendiculars: this is a known fact from the orthology of \( ABC \) and \( PQR \).
- 0 points for \( H, Y, Z, D \) cyclic.
- 1 point for proving \( A'B'C'D'E'F' \) are collinear; no points for just conjecturing this.
- 1 point for realizing that \( A'D' \) is the radical axis of the two relevant circles, even without proof.
- 1 point for realizing that the concurrence point is the reflection of \( H \) across the center of \( (XYZ) \), even without proof.
- 2 points for noticing \( D'QF' \) and \( ACO \) are orthologic (or analogous).
|
USEMO-2020-1
|
https://web.evanchen.cc/exams/report-usemo-2020.pdf
|
Which positive integers can be written in the form
\[
rac{\operatorname{lcm}(x, y) + \operatorname{lcm}(y, z)}{\operatorname{lcm}(x, z)}
\]
for positive integers \(x, y, z\)?
|
Let \(k\) be the desired value, meaning
\[
-k \operatorname{lcm}(x, z) + \operatorname{lcm}(x, y) + \operatorname{lcm}(y, z) = 0.
\]
Our claim is that the possible values are even integers.
Indeed, if \(k\) is even, it is enough to take \((x, y, z) = (1, k/2, 1)\).
For the converse direction we present a few approaches.
**Solution 1: First approach using \(
u_2\) only.**
We are going to use the following fact:
**Lemma.** If \(u, v, w\) are nonzero integers with \(u + v + w = 0\), then either
\[
egin{aligned}
u_2(u) &>
u_2(v) =
u_2(w); \
u_2(v) &>
u_2(w) =
u_2(u); \quad ext{or} \
u_2(w) &>
u_2(u) =
u_2(v).
\end{aligned}
\]
*Proof.* Letâs assume WLOG that \(e =
u_2(w)\) is minimal. If both \(
u_2(u)\) and \(
u_2(v)\) are strictly greater than \(e\), then \(
u_2(u + v + w) = e\) which is impossible. So assume WLOG again that \(
u_2(v) =
u_2(w) = e\). Then
\[
u_2(u) \ge e + 1.\]
However, if we assume for contradiction that \(k\) is odd, then
\[
egin{aligned}
u_2(-k \operatorname{lcm}(x, z)) &= \max(
u_2(x),
u_2(z)) \
u_2(\operatorname{lcm}(x, y)) &= \max(
u_2(x),
u_2(y)) \
u_2(\operatorname{lcm}(y, z)) &= \max(
u_2(y),
u_2(z)).
\end{aligned}
\]
In particular, the *largest* two numbers among the three right-hand sides must be equal. So by the lemma, there is no way the three numbers \(-k \operatorname{lcm}(x, z), \operatorname{lcm}(x, y), \operatorname{lcm}(y, z)\) could have sum zero.
**Solution 2: Second approach using \(
u_p\) for general \(p\).**
Weâll prove the following much stronger claim (which will obviously imply \(k\) is even).
**Claim.** We must have \(\operatorname{lcm}(x, z) \mid \operatorname{lcm}(x, y) = \operatorname{lcm}(y, z)\).
*Proof.* Take any prime \(p\) and look at three numbers \(
u_p(x),
u_p(y),
u_p(z)\). Weâll show that
\[
\max(
u_p(x),
u_p(z)) \le \max(
u_p(x),
u_p(y)) = \max(
u_p(y),
u_p(z)).
\]
If \(
u_p(y)\) is the (non-strict) maximum, then the claim is obviously true.
If not, by symmetry assume WLOG that \(
u_p(x)\) is largest, so that \(
u_p(x) >
u_p(y)\) and \(
u_p(x) \ge
u_p(z)\). However, from the given equation, we now have \(
u_p(\operatorname{lcm}(y, z)) \ge
u_p(x)\). This can only occur if \(
u_p(z) =
u_p(x)\). So the claim is true in this case too.
**Solution 3: Third approach without taking primes (by circlethm).**
By scaling, we may as well assume \(\gcd(x, y, z) = 1\).
Let \(d_{xy} = \gcd(x, y)\), etc. Now note that \(\gcd(d_{xy}, d_{xz}) = 1\), and cyclically! This allows us to write the following decomposition:
\[
egin{aligned}
x &= d_{xy} d_{xz} a \
y &= d_{xy} d_{yz} b \
z &= d_{xz} d_{yz} c
\end{aligned}
\]
We also have \(\gcd(a, b) = \gcd(b, c) = \gcd(c, a) = 1\) now.
Now, we have
\[
egin{aligned}
\operatorname{lcm}(x, y) &= d_{xy} d_{xz} d_{yz} ab \
\operatorname{lcm}(y, z) &= d_{xy} d_{xz} d_{yz} bc \
\operatorname{lcm}(x, z) &= d_{xy} d_{xz} d_{yz} ac
\end{aligned}
\]
and so substituting this in to the equation gives
\[
k = b \cdot \left( rac{1}{a} + rac{1}{c}
ight).
\]
For \(a, b, c\) coprime this can only be an integer if \(a = c\), so \(k = 2b\).
**Remark.** From \(a = c = 1\), the third approach also gets the nice result that \(\operatorname{lcm}(x, y) = \operatorname{lcm}(y, z)\) in the original equation.
|
νâ solutions
⢠0 points for stating that even integers are the only solutions
⢠1 point for the construction for even integers
⢠1 points for considering νâ of x, y and z and resolving at least one substantial case, such as νâ(y) being maximal (additive). This point can also be awarded if done with a general prime p instead of 2.
⢠7 points for a complete solution
Factoring solutions
⢠0 points for stating that even integers are the only solutions
⢠0 points for stating WLOG gcd(x, y, z) = 1
⢠1 point for the construction for even integers
⢠1 points for factoring pairwise greatest common divisors (additive)
⢠7 points for a complete solution
General νâ solutions
⢠0 points for stating that even integers are the only solutions
⢠1 point for the construction for even integers
⢠1 point for stating lcm(x, z) | lcm(x, y) = lcm(y, z) (additive)
⢠7 points for a complete solution
|
USEMO-2020-2
|
https://web.evanchen.cc/exams/report-usemo-2020.pdf
|
Calvin and Hobbes play a game. First, Hobbes picks a family \( \mathcal{F} \) of subsets of \( \{1, 2, \ldots, 2020\} \), known to both players. Then, Calvin and Hobbes take turns choosing a number from \( \{1, 2, \ldots, 2020\} \) which is not already chosen, with Calvin going first, until all numbers are taken (i.e., each player has 1010 numbers). Calvin wins if he has chosen all the elements of some member of \( \mathcal{F} \), otherwise Hobbes wins. What is the largest possible size of a family \( \mathcal{F} \) that Hobbes could pick while still having a winning strategy?
|
The answer is \( 4^{1010} - 3^{1010} \). In general, if 2020 is replaced by \( 2n \), the answer is \( 4^n - 3^n \).
**Construction.** The construction is obtained as follows: pair up the numbers as \( \{1,2\}, \{3,4\}, \ldots, \{2019,2020\} \). Whenever Calvin picks a number from one pair, Hobbes elects to pick the other number. Then Calvin can never obtain a subset which has both numbers from one pair. There are indeed \( 2^{2n} - 3^n \) subsets with this property, so this maximum is achieved.
**Bound.** The main claim is the following.
**Claim â** Fix a strategy for Hobbes and an integer \( 0 \le k \le n \). Then there are at least \( inom{n}{k} 2^k \) sets with \( k \) numbers that Calvin can obtain after his \( k \)th turn.
*Proof, due to Andrew Gu.* The number of ways that Calvin can choose his first \( k \) moves is
\[
2n \cdot (2n - 2) \cdot (2n - 4) \cdots (2n - 2(k - 1)).
\]
But each \( k \)-element set can be obtained in this way in at most \( k! \) ways (based on what order its numbers were taken). So we get a lower bound of
\[
rac{2n \cdot (2n - 2) \cdot (2n - 4) \cdots (2n - 2(k - 1))}{k!} = 2^k inom{n}{k}.
\]
Thus by summing \( k = 0, \ldots, n \) the family \( S \) is missing at least
\[
\sum_{k=0}^n 2^k inom{n}{k} = (1+2)^n = 3^n
\]
subsets, as desired.
**Alternate proof of bound.** Fix a strategy for Hobbes, as before. We proceed by induction on \( n \) to show there are at least \( 3^n \) missing sets (where a âmissing setâ, like in the previous proof, is a set that Calvin can necessarily reach). Suppose that if Calvin picks 1 then Hobbes picks 2. Then the induction hypothesis on the remaining game gives that:
- there are \( 3^{n-1} \) missing sets that contain 1 but not 2;
- there are also \( 3^{n-1} \) missing sets that contain neither 1 nor 2;
- But imagining Calvin picking 2 first instead, applying the induction hypothesis again we find that there are \( 3^{n-1} \) missing sets which contain 2.
These categories are mutually exclusive, so we find there are at least \( 3^n \) missing sets, as needed.
|
- **1 point** for a correct construction, with the correct strategy for Hobbes which results in the correct answer (but they need not correctly calculate the answer, as long as the strategy is right).
- **2 points** for claiming a statement (no proof needed) of the form âfor each integer \( k \), Calvin can ensure a victory if the number of sets of size \( k \) is at most ââ (or âthe number of missing sets is at least ââ) and the bound listed is the correct (tight) bound.
- **3 points** for both of the above items.
- **5 points** for a correct proof of the upper bound.
- **7 points** for a fully correct solution.
- **-2 points** for solutions which would satisfy rubric items (3) or (4), except that the expressions given in terms of \( n \) and \( k \) are missing or not bounded correctly.
For example: a solution might have a claim that Hobbes needs to be missing some number of sets of size \( k \), and show how to use induction to show a bound on \( n, k \) based on \( n-1, k-1 \), but not use the correct quantities to do the bounding.
- **-1 point** for a wrong answer, only for solutions that would otherwise receive a 7. Deduct no points if an answer is unsimplified but correct.
Deductions for other minor/major errors are as usual. None of the items above are additive (though see the third rubric item).
In general, for solutions that do not fit a rubric item but which you feel deserves partial credit, use your judgment to assign partial credit.
|
USEMO-2020-3
|
https://web.evanchen.cc/exams/report-usemo-2020.pdf
|
Let \( ABC \) be an acute triangle with circumcenter \( O \) and orthocenter \( H \). Let \( \Gamma \) denote the circumcircle of triangle \( ABC \), and \( N \) the midpoint of \( \overline{OH} \). The tangents to \( \Gamma \) at \( B \) and \( C \), and the line through \( H \) perpendicular to line \( AN \), determine a triangle whose circumcircle we denote by \( \omega_A \). Define \( \omega_B \) and \( \omega_C \) similarly.
Prove that the common chords of \( \omega_A, \omega_B, \omega_C \) are concurrent on line \( OH \).
|
We begin by introducing several notations. The orthic triangle is denoted \( DEF \) and the tangential triangle is denoted \( T_aT_bT_c \). The reflections of \( H \) across the sides are denoted \( H_a, H_b, H_c \). We also define the crucial points \( P \) and \( Q \) as the poles of \( H_cB \) and \( H_bC \) with respect to \( \Gamma \).
The solution, based on the independent solutions found by Anant Mudgal and Nikolai Beluhov, hinges on two central claims: that \( \omega_A \) is the circumcircle of \( riangle T_aPQ \), and that \( \overline{EF} \) is the radical axis of \( \Gamma \) and \( \omega_A \). We prove these two claims in turn.
**Claim (Characterization of \( \omega_A \)) â Line \( PQ \) passes through \( H \) and is perpendicular to \( \overline{AN} \).**
*Proof.* The fact that \( H \) lies on line \( PQ \) is immediate by Brokardâs theorem.
Showing the perpendicularity is the main part. Denote by \( B' \) and \( C' \) the antipodes of \( B \) and \( C \) on \( \Gamma \). Also, define \( L = \overline{H_cC'} \cap \overline{H_bB'} \) and \( K = \overline{BH_c} \cap \overline{CH_b} \).
We observe that:
- We have \( \overline{OK} \perp \overline{PQ} \) since \( K \) is the pole of line \( \overline{PQ} \) (again by Brokard).
- The points \( O, K, L \) are collinear by Pascalâs theorem on \( BH_cC'/CH_bB' \).
- The point \( L \) is seen to be the reflection of \( H \) across \( A \), so it follows \( \overline{AN} \parallel \overline{OL} \) by a \( frac{1}{2} \)-factor homothety at \( H \).
Putting these three observations together completes the first claim. \( \square \)
**Remark.** (First claim is faster with complex numbers.) It is also straightforward to prove the first claim by using complex numbers. Indeed, in the usual setup, we have that the intersection of the tangents at \( B \) and \( H_c \) is given explicitly by
\[
p = rac{2b \cdot \left(- rac{ab}{c}
ight)}{b - rac{ab}{c}} = rac{2ab}{a - c}
\]
and one explicitly checks \( p - (a + b + c) \perp (b + c - a) \), as needed.
**Claim (Radical axis of \( \omega_A \) and \( \Gamma \)) â Line \( EF \) coincides with the radical axis of \( \omega_A \) and \( \Gamma \).**
*Proof.* Let lines \( EF \) and \( T_aT_c \) meet at \( Z \). It suffices to show \( Z \) lies on the radical axis, and then repeat the argument on the other side.
Since \( ngle FBZ = ngle ABZ = ngle BCA = ngle EFA = ngle ZFB \), it follows \( ZB = ZF \). We introduce two other points \( X \) and \( Y \) on the perpendicular bisector of \( BF \): they are the midpoints of \( BC \) and \( BH_c \).
Since \( OX \cdot OT_a = OB^2 = OY \cdot OP \), it follows that \( XYPT_a \) is cyclic. Then
\[
ZP \cdot ZT_a = ZX \cdot ZY = ZB^2
\]
with the last equality since the circumcircle of \( riangle BXY \) is tangent to \( \Gamma \) (by a \( frac{1}{2} \)-homothety at \( B \)). So the proof of the claim is complete. \( \square \)
**Claim â Line \( DT_a \) coincides with the radical axis of \( \omega_B \) and \( \omega_C \).**
*Proof.* The point \( D \) already coincides with the radical axis because it is the radical center of \( \Gamma, \omega_B \) and \( \omega_C \). As for the point \( T_a \), we let the tangent to \( \Gamma \) at \( H_a \) meet \( T_aT_c \) at \( U \) and \( V \); by the first claim, these lie on \( \omega_C \) and \( \omega_B \) respectively.
We need to show \( T_aU \cdot T_aT_c = T_aV \cdot T_aT_b \).
But \( UVT_bT_c \) is apparently cyclic: the sides \( T_bT_c \) and \( UV \) are reflections across a line perpendicular to \( AH_a \), while the sides \( UT_c \) and \( VT_b \) are reflections across a line perpendicular to \( BC \). So this is true. \( \square \)
Now since \( riangle DEF \) and \( riangle T_aT_bT_c \) are homothetic (their opposite sides are parallel), and their incenters are respectively \( H \) and \( O \), the problem is solved.
**Remark.** (Barycentric approaches with respect to \( riangle T_aT_bT_c \)). Because the first claim is so explicit, it is possible to calculate the length of the segment \( PB \). This opens the possibility of using barycentric coordinates with respect to the reference triangle \( T_aT_bT_c \), and in fact some contestants were able to complete this approach. Writing \( a = T_bT_c, b = T_cT_a, c = T_aT_b \) one can show that the radical center is the point
\[
\left( rac{a}{s - a} : rac{b}{s - b} : rac{c}{s - c}
ight)
\]
which is checked to be collinear with the circumcenter and incenter of \( riangle T_aT_bT_c \).
---
**Alternate inversion approach replacing the last two claims, by Serena An.**
After finding \( P \) and \( Q \), itâs also possible to solve the problem by using inversion. This eliminates the need to identify line \( EF \) as the radical axis of \( \omega_A \) and \( \Gamma \).
Inverted points are denoted with \( ullet^* \) as usual, but we will only need two points: \( P^* \), the midpoint of \( BH_c \), and \( Q^* \), the midpoint of \( CH_b \). Now, let \( M_a = T_a^* \) denote the midpoint of \( BC \) and let \( K \) be a point on ray \( HA \) with
\[
KH = rac{3}{2} AH.
\]
**Claim â The points \( D, P^*, Q^* \) lie on the circle with diameter \( KM_a \).**
*Proof.* Consider the homothety at \( H \) with scale factor \( frac{3}{2} \). It maps \( F \) to the midpoint of \( FH_c \) and \( A \) to \( K \), so we find \( KP^* \) is the perpendicular bisector of \( FH_c \). As \( M_aP^* \parallel CH \), we conclude \( ngle KP^*M_a = 90^\circ \).
Similarly \( ngle KQ^*M_a = 90^\circ \). And \( ngle KDM_a = 90^\circ \) is given. \( \square \)
**Claim â Line \( CH \) coincides with the radical axis of \( \omega_A^* \) and \( \omega_B^* \). In particular, the circles \( \omega_A, \omega_B, (COH_c) \) are coaxial.**
*Proof.* Letting \( \Gamma \) and \( \Gamma_9 \) denote circumcircle and nine-point circle,
\[
ext{Pow}(C, \omega_A^*) = CD \cdot CM_a = ext{Pow}(C, \Gamma_9) \
ext{Pow}(H, \omega_A^*) = HM \cdot HD = frac{3}{2} HA \cdot frac{1}{2} HH_a = frac{3}{4} ext{Pow}(H, \Gamma).
\]
The same calculation holds with \( \omega_B^* \). Now line \( CH \) inverts to \( (COH_c) \), as needed. \( \square \)
Since the circles \( (AOH_a), (BOH_b), (COH_c) \) have common radical axis equal to line \( OH \), the problem is solved.
---
**Remark.** (Nikolai Beluhov â generalization with variable \( ABC \) and fixed \( H \)). Take a fixed circle \( \Gamma \) and a fixed point \( H \) in its interior. Then there exist infinitely many triangles \( ABC \) with orthocenter \( H \) and circumcircle \( \Gamma \). In fact, for every point \( A \) on \( \Gamma \) we get a unique pair of \( B \) and \( C \), determined as follows: Let line \( AH \) meet \( \Gamma \) again at \( S_A \), and take \( B \) and \( C \) to be the intersection points of the perpendicular bisector of segment \( HS_A \) with \( \Gamma \).
With this framework, the following generalization is true: The radical center \( W \) of \( \omega_A, \omega_B, \omega_C \) is the same point for all such triangles. Indeed, the Euler circle of triangle \( ABC \) is constant because it depends only on \( H \) and \( \Gamma \). Let inversion relative to \( \Gamma \) map \( e \) onto \( \Omega \). Then all three of \( T_a, T_b, T_c \) lie on \( \Omega \), and so, by the solution, \( W \) is the homothety center of \( e \) and \( \Omega \). (We take the homothety center with positive ratio when triangle \( ABC \) is acute, and with negative ratio when it is obtuse. When triangle \( ABC \) is right-angled, \( \Omega \) degenerates into a straight line.)
Explicitly, let \( \Gamma \) be the unit circle and put \( H \) on the real axis at \( h \). Then \( W \) is also on the real axis, at \( 4h/(h^2 + 3) \).
Furthermore, it turns out the power of \( W \) with respect to \( \omega_A, \omega_B, \omega_C \) is constant as well. This, however, is much tougher to prove; we are not aware of a purely geometric proof at this time. Explicitly, in the setting above where \( \Gamma \) is the unit circle and \( H \) is on the real axis at \( h \), the power of \( W \) with respect to \( \omega_A, \omega_B, \omega_C \) equals \( 12(h^2 - 1)/(h^2 + 3)^2 \).
|
As usual, incomplete computational approaches earn partial credits only based on the amount of synthetic progress which is made.
No points are awarded for just drawing a diagram or simple observations.
Follow the notation in the typeset official solution. The following rubric items are totally additive:
(a) **1 point** for proving that \( P \) and \( Q \) are the poles of lines \( BH_c \) and \( CH_b \).
(b) **1 point** for proving that \( T_a \) is on the radical axis of \( \omega_B \) and \( \omega_C \). This point can be awarded if the proof is conditional on some reasonable description of \( P \) and \( Q \), such as (a).
(c) **2 points** for proving that \( D \) is on the radical axis of \( \omega_B \) and \( \omega_C \). This point can be awarded if the proof is conditional on some reasonable description of \( P \) and \( Q \), such as (a).
(d) **0 points** for commenting that the homothety center of \( T_aT_bT_c \) and \( DEF \) lies on the line \( OH \).
The four rubric items above, when combined, give a perfect solution worth 7.
If none of the items above are earned: the following rubric item (not additive) is possible:
- **1 point** for both claiming that \( P \) and \( Q \) are the poles of \( BH_c \) and \( CH_b \), and that the radical axis of \( \omega_B \) and \( \omega_C \) is exactly \( DT_a \).
|
USEMO-2020-4
|
https://web.evanchen.cc/exams/report-usemo-2020.pdf
|
A function \( f \) from the set of positive real numbers to itself satisfies
\[
f(x + f(y) + xy) = xf(y) + f(x + y)
\]
for all positive real numbers \( x \) and \( y \). Prove that \( f(x) = x \) for all positive real numbers \( x \).
|
Solution 1
We first begin with the following observation.
**Claim** â We must have \( f(y) \ge y \) for all \( y > 0 \).
*Proof.* Otherwise, choose \( 0 < x < 1 \) satisfying that \( f(y) = (1 - x) \cdot y \). Then plugging this \( P(x, y) \) gives \( xf(y) = 0 \), contradiction. \(\square\)
Now, we make the substitution \( f(x) = x + g(x) \), so that \( g \) is a function \( \mathbb{R}_{>0} o \mathbb{R}_{\ge 0} \). The given function equation reads
\[
g(x + xy + (y + g(y))) + x + (y + g(y)) = (xy + xg(y)) + (x + y + g(x + y)),
\]
or
\[
g(x + y + xy + g(y)) = (x - 1)g(y) + g(x + y). ag{\dagger}
\]
We have to show that \( g \) is the zero function from \( (\dagger) \).
**Claim (Injectivity for nonzero outputs)** â If \( g(a) = g(b) \) for \( a
e b \), then we must actually have \( g(a) = g(b) = 0 \).
*Proof.* Setting \( (a, b) \) and \( (b, a) \) in \( (\dagger) \) gives
\[
(a - 1)g(b) = (b - 1)g(a)
\]
which, since \( a - 1
e b - 1 \), forces \( g(a) = g(b) = 0 \). \(\square\)
**Claim (\( g \) vanishes on \( (1, \infty) \))** â We have \( g(t) = 0 \) for \( t > 1 \).
*Proof.* If we set \( x = 1 \) in \( (\dagger) \) we obtain that
\[
g(g(y) + 2y + 1) = g(1 + y).
\]
As the inputs are obviously unequal, the previous claim gives \( g(1 + y) = 0 \) for all \( y > 0 \). \(\square\)
Now \( x = 2 \) in \( (\dagger) \) to get \( g(y) = 0 \), as needed.
Solution 2
We start with the same opening of showing \( f(y) \ge y \), defining \( f(x) = x + g(x) \), so \( g \) satisfies \( (\dagger) \). Here is another proof that \( g \equiv 0 \) from \( (\dagger) \).
**Claim** â If \( g \) is not the zero function, then for any constant \( C \), we have \( g(t) > C \) for sufficiently large \( t \).
*Proof.* In \( (\dagger) \) fix \( y \) to be any input for which \( g(y) > 0 \). Then
\[
g\left((1 + y)x + (y + g(y))
ight) \ge (x - 1)g(y)
\]
so for large \( x \), we get the conclusion. \(\square\)
On the other hand, by choosing \( x = 1 \) and \( y = t - 1 \) for \( t > 1 \) in \( (\dagger) \), we get
\[
g(2t + g(z) - 1) = g(t)
\]
and hence one can generate an infinite sequence of fixed points: start from \( t_0 = 100 \), and define
\[
t_n = 2t_{n-1} + g(t_{n-1}) - 2 > t_{n-1} + 98
\]
for \( n \ge 1 \) to get
\[
g(t_0) = g(t_1) = g(t_2) = \cdots
\]
and since the \( t_i \) are arbitrarily large, this produces a contradiction if \( g
e 0 \).
|
General remarks
- Unlike most functional equations, this one doesnât really have that many steps of âpartial progress.â As such, the steps below are mainly intermediate claims, not specific equations.
- This problem does not ask the contestant to find all functions \( f \) that satisfy the given property. As a result, the fact that \( f(x) = x \) satisfies the property does not need to be stated nor proven.
- The results below may come in many different forms, and may not be stated explicitly. As such, care must be taken to determine whether a contestant has found any of the claims below.
- Most solutions follow the following general path:
(a) Show \( f(x) \ge x \), and define \( g(x) = f(x) - x \).
(b) Find some property that shows \( g(x) \) must be small or fixed in some places.
(c) Find some property that shows \( g(x) \) must be large in some places.
(d) Combine the two to finish.
The first claim is worth 1 point, and points (b) and (c) are meant to each be worth 3 points (non-additively), with a full solution (including finish) worth 7, of course. Solutions that make enough progress that a âstandard finishâ is applicable should be worth 5 points (in essence, points for both (b)-like and (c)-like things should be given). The rubric below attempts to codify most (b)-like and (c)-like progress we expect to see, but it is certainly possible that other progress exists.
Rubric items
None of these items are additive.
- 0 points for solving the equation over some domain that is not the positive reals (e.g. reals, nonnegative reals)
- 1 point for showing \( f(x) \ge x \) for all \( x \).
- 1 point for showing that, if \( f(x) = x \) for some \( x \), then \( f(x) = x \) for all \( x \).
- 3 points for showing that, if \( f(x) + y = f(y) + x \), then \( f(x) = x \) and \( f(y) = y \).
- 3 points for finding distinct values of \( x \) and \( y \) for which \( f(x) + y = f(y) + x \).
- 3 points for showing that \( f(x) - x \) is eventually greater than any specified real.
- 5 points for finding (or showing existence of) any value of \( x \) for which \( f(x) = x \).
- 6 points for a complete solution with a minor error that does not affect the solution.
- 7 points for a complete solution.
|
USEMO-2020-5
|
https://web.evanchen.cc/exams/report-usemo-2020.pdf
|
The sides of a convex 200-gon \( A_1A_2 \ldots A_{200} \) are colored red and blue in an alternating fashion. Suppose the extensions of the red sides determine a regular 100-gon, as do the extensions of the blue sides.
Prove that the 50 diagonals \( A_1A_{101},\ A_3A_{103},\ \ldots,\ A_{99}A_{199} \) are concurrent.
|
Let \( B_1 \ldots B_{100} \) and \( R_1 \ldots R_{100} \) be the regular 100-gons (oriented counterclockwise), and define \( X_i = A_{2i+1} = \overline{B_iB_{i+1}} \cap \overline{R_iR_{i+1}} \) for all \( i \), where all indices are taken modulo 100.
We wish to show that \( X_1X_{51}, \ldots, X_{50}X_{100} \) are concurrent.
We now present two approaches.
**Solution 1**
*First approach (by spiral similarity).* Let \( O \) be the spiral center taking \( B_1 \ldots B_{100} o R_1 \ldots R_{100} \) (it exists since the 100-gons are not homothetic). We claim that \( O \) is the desired concurrency point.
**Claim** â \( ngle X_iOX_{i+1} = rac{\pi}{50} \) for all \( i \).
*Proof.* Since \( riangle OB_iB_{i+1} \sim riangle OR_iR_{i+1} \), we have \( riangle OB_iR_i \sim riangle OB_{i+1}R_{i+1} \), so \( O, X_i, B_{i+1}, R_{i+1} \) are concyclic. Similarly \( O, X_{i+1}, B_{i+1}, R_{i+1} \) are concyclic, so
\[ ngle X_iOX_{i+1} = ngle X_iB_{i+1}X_{i+1} = rac{\pi}{50} \]
as wanted. \( \square \)
It immediately follows that \( O \) lies on all 50 diagonals \( X_iX_{i+50} \), as desired.
**Solution 2**
*Second approach (by complex numbers).* Let \( \omega \) be a primitive 100th root of unity. We will impose complex coordinates so that \( R_k = \omega^k \), while \( B_k = p\omega^k + q \), where \( m \) and \( b \) are given constant complex numbers.
In general for \( |z| = 1 \), we will define \( f(z) \) as the intersection of the line through \( z \) and \( \omega z \), and the line through \( pz + q \) and \( p \cdot \omega z + q \).
In particular, \( X_k \) is \( f(\omega^k) \).
**Claim** â There exist complex numbers \( a, b, c \) such that \( f(z) = a + bz + cz^2 \), for every \( |z| = 1 \).
*Proof.* Since \( f(z) \) and \( rac{f(z)-q}{p} \) both lie on the chord joining \( z \) to \( \omega z \) we have
\[ z + \omega z = f(z) + \omega z^2 \cdot \overline{f(z)} \]
\[ z + \omega z = rac{f(z) - q}{p} + \omega z^2 \cdot rac{\overline{f(z)} - \overline{q}}{\overline{p}}. \]
Subtracting the first equation from the \( \overline{p} \) times the second to eliminate \( \overline{f(z)} \), we get that \( f(z) \) should be a degree-two polynomial in \( z \) (where \( p \) and \( q \) are fixed constants). \( \square \)
**Claim** â Let \( f(z) = a + bz + cz^2 \) as before. Then the locus of lines through \( f(z) \) and \( f(-z) \), as \( |z| = 1 \) varies, passes through a fixed point.
*Proof.* By shifting we may assume \( a = 0 \), and by scaling we may assume \( b \) is real (i.e. \( \overline{b} = b \)). Then the point \( -\overline{c} \) works, since
\[ rac{f(z) + \overline{c}}{f(-z) + \overline{c}} = rac{\overline{c} + bz + cz^2}{\overline{c} - bz + cz^2} \]
is real â it obviously equals its own conjugate. (Alternatively, without the assumptions \( a = 0 \) and \( b \in \mathbb{R} \), the fixed point is \( a - rac{\overline{b}c}{b} \)). \( \square \)
**Remark** (We know a priori the exact coefficients shouldnât matter). In fact, the exact value is
\[ f(z) = rac{-\omega \overline{q} z^2 + (1 - \overline{p})(1 + \omega)z - rac{\overline{p}}{p}q}{1 - rac{\overline{p}}{p}}. \]
Since \( p \) and \( q \) could be any complex numbers, the quantity \( c/b \) (which is all that matters for concurrence) could be made to be equal to any value. For this reason, we know a priori the exact coefficients should be irrelevant.
|
Most solutions are worth 0 or 7.
- 0 points for no progress, special cases, etc.
- 5â6 points for any tiny slip which the contestant could have easily repaired
- 7 points for a correct solution
For solutions which are not complete, the following items are additive:
- 1 point for considering the spiral similarity taking \( P_1 \ldots P_{100} \) to \( Q_1 \ldots Q_{100} \) AND claiming that the center of the spiral similarity is the point of concurrency.
- 1 point for claiming that \( ngle R_iOR_{i+1} = rac{\pi}{50} \)
- 1 point for proving that \( O, R_i, P_{i+1}, Q_{i+1} \) is concyclic
- 1 point for further extending the above to proving that \( O, R_i, R_{i+1}, P_{i+1}, Q_{i+1} \) concyclic
There is no deduction for small configuration issues (such as not using directed angles) or small typos (such as labelling points).
Usually, computational approaches which are not essentially completed are judged by any geometric content and do not earn other marks. However, the following marks (not additive with anything) are possible:
- Following the notation of the complex solution, 1 point for showing that the intersection point is quadratic AND making the general claim that \( a + bz + cz^2 \) is sufficient regardless of what the numbers \( a, b, c \) are.
|
USEMO-2020-6
|
https://web.evanchen.cc/exams/report-usemo-2020.pdf
|
Prove that for every odd integer \( n > 1 \), there exist integers \( a, b > 0 \) such that, if we let \[ Q(x) = (x + a)^2 + b, \] then the following conditions hold:
- we have \( \gcd(a,n) = \gcd(b,n) = 1 \);
- the number \( Q(0) \) is divisible by \( n \); and
- the numbers \( Q(1), Q(2), Q(3), \ldots \) each have a prime factor not dividing \( n \).
|
Let \( p_1 < p_2 < \cdots < p_m \) denote the odd primes dividing \( n \) and call these primes small. The construction is based on the following idea:
**Claim** â For each \( i = 1, \ldots, m \), we can choose a prime \( q_i \equiv 1 \pmod{4} \) such that
\[
\left( \frac{p_j}{q_i} \right) = \begin{cases}
-1 & \text{if } j = i \
+1 & \text{otherwise.}
\end{cases}
\]
*Proof.* Fix \( i \). By quadratic reciprocity, it suffices that \( q_i \equiv 1 \pmod{4} \) and that \( q_i \) is a certain nonzero quadratic residue (or not) modulo \( p_j \) for \( j \ne i \).
By Chinese remainder theorem, this is a single modular condition, so Dirichlet theorem implies such primes exist. \( \square \)
We commit now to the choice
\[
b = kq_1q_2 \cdots q_m
\]
where \( k \ge 1 \) is an integer (its value does not affect the following claim).
**Claim (Main argument)** â For this \( b \), there are only finitely many integers \( X \) satisfying the equation
\[
X^2 + b = p_1^{e_1} \cdots p_m^{e_m} \tag{\spadesuit}
\]
where \( e_i \) are some nonnegative integers (i.e. \( X^2 + b \) has only small prime factors).
*Proof.* In (\( \spadesuit \)) the RHS is a quadratic residue modulo \( b \). For any \( i > 0 \), modulo \( q_i \) we find
\[
+1 = \prod_j \left( \frac{p_j^{e_j}}{q_i} \right) = (-1)^{e_i}
\]
so \( e_i \) must be even. This holds for every \( i \) though! In other words all \( e_i \) are even.
Hence (\( \spadesuit \)) gives solutions to \( X^2 + b = Y^2 \), which obviously has only finitely many solutions. \( \square \)
We now commit to choosing any \( k \ge 1 \) such that
\[
k \equiv -\frac{1}{q_1q_2 \cdots q_m} \pmod{n}
\]
which in particular means \( \gcd(k,n) = 1 \). Now as long as \( a \equiv 1 \pmod{n} \), we have \( Q(0) \equiv 0 \pmod{n} \), as needed. All that remains is to take \( a \) satisfying the second claim larger than any of the finitely many bad integers in the first claim.
|
In general, not much partial credit is expected for this problem.
The heart of the problem can be thought of as studying the equation
\[
X^2 + b = p_1^{e_1} \cdots p_k^{e_k}
\]
where \( p_1, \ldots, p_k \) are a fixed set of primes, and showing that the equation cannot hold for all sufficiently large \( X \).
- **No points** are given for steps related to the first two conditions, e.g. for the Henselâs-lemma type observation that \( e_i \) may be unbounded. This is equivalent to reducing to the case where \( n \) is squarefree.
- **No points** are given for taking specific moduli, e.g. taking the equation modulo 4.
- **No points** are given for special cases of \( k \), such as \( k = 1 \).
- **1 point** is awarded for the idea to select a prime \( q \) for which \( \left( \frac{p_i}{q} \right) \) is known in order to control the parity of the exponents \( e_i \).
- There is no deduction for quoting the theorems of Dirichlet or quadratic reciprocity, or in general the quoting of any named theorems which the grader can indeed verify exists.
- **1 point is deducted** if the student fails to verify the Hensel argument, but their construction holds anyways.
- **1 point is deducted** if the student asserts control over \( \left( \frac{p_i}{q} \right) \) with no justification whatsoever. We require the student to at least mention that quadratic reciprocity is used to get a modular condition and then use Dirichlet. (This justification may be very terse: even âby QR and Dirichletâ is accepted).
|
USEMO-2021-1
|
https://web.evanchen.cc/exams/report-usemo-2021.pdf
|
Let \( n \) be a positive integer and consider an \( n imes n \) grid of real numbers. Determine the greatest possible number of cells \( c \) in the grid such that the entry in \( c \) is both strictly greater than the average of \( c \)'s column and strictly less than the average of \( c \)'s row.
|
The answer is \((n-1)^2\). An example is given by the following construction, shown for \(n = 5\), which generalizes readily. Here, the lower-left \((n-1) imes (n-1)\) square gives a bound.
\[
egin{bmatrix}
-1 & -1 & -1 & -1 & 0 \
0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\]
We give two proofs of the bound. Call a cell \emph{good} if it satisfies the condition of the problem.
Solution 1
extbf{Coloring proof, from the author.} We now prove that no more than \((n-1)^2\) squares can be good. The cells are weakly ordered by \( \ge \) (there may be some ties due to equal elements); we arbitrarily extend it to a total ordering \( > \), breaking all ties. (Alternatively, one can phrase this as perturbing the grid entries in such a way that they become distinct.)
- For every column, we color red the \(>\)-smallest element in that column.
- For every row, we color blue the \(>\)-largest element in that row.
This means that there are exactly \(n\) red and \(n\) blue cells. Note that these cells are never good.
extbf{Claim} â There is at most one cell that is both red and blue.
\emph{Proof.} Assume for contradiction that \(P_1\) and \(P_2\) are two âpurpleâ cells (both red and blue). Look at the resulting picture
\[
egin{bmatrix}
P_1 & x \
y & P_2
\end{bmatrix}.
\]
By construction, we have \(P_1 > x > P_2 > y > P_1\). This is a contradiction. \(\square\)
Thus at least \(2n - 1\) cells cannot be good. This proves the bound.
Solution 2
extbf{Proof using KÅnigâs theorem, from Ankan Bhattacharya.} This proof is based on the following additional claim:
extbf{Claim} â No column/row can be all-good, and no transversal can be all-good.
\emph{Proof.} The first part is obvious. As for the second, let \(r_i\) and \(c_j\) denote the column sums. If cell \((i,j)\) is good, then
\[ r_i < a_{i,j} < c_j. \]
If we have a good transversal, summing the inequality \(r_i < c_j\) over the cells in this transversal gives a contradiction (as \( \sum \mathbf{r}_ullet = \sum \mathbf{c}_ullet \)). \(\square\)
This claim alone is enough to imply the desired bound.
extbf{Claim} â There exists a choice of \(a\) columns and \(b\) rows, with \(a + b = n + 1\), such that no good cells lie on the intersection of the columns and rows.
\emph{Proof.} Follows by \emph{KÅnigâs theorem}, and the previous claim. Alternatively, quote the contrapositive of Hallâs marriage theorem: because there was no all-good transversal, there must be a set of \(a\) columns with more than \(n - a\) âcompatibleâ rows. \(\square\)
Suppose \(a \le rac{n}{2} \le b\); the other case is similar. Now we bound:
- The \(a imes b\) cells of the claim are given to be all non-good.
- In the \(n - a = b - 1\) remaining rows, there is at least one more non-good cell.
Thus the number of non-good cells is at least
\[
ab + (b - 1) = (a + 1)b - 1 \ge 2 \cdot n - 1 = n^2 - (n - 1)^2,
\]
and so there are at most \((n - 1)^2\) good cells.
|
In all approaches, extbf{1 point} is awarded for a correct construction, and extbf{6 points} for the proof that \((n - 1)^2\) is best possible; hence \(1 + 6 = 7\). No points are awarded for answer alone.
The 6 points can be decomposed according to the following approaches. All the items, including the deductions, are additive within each approach. But the marks from different approaches are not additive.
Finally, extbf{one point is deducted} from a 7â solution if the proof is only correct when all the cells are distinct, and the student fails to deal with situations in which a subset of cells have the same number. (This is quite forgiving.)
extbf{First approach}
- 1 point for the claim that there exists only 1 cell which is the largest and smallest in its row and column respectively
- 5 points for the proof and by showing that otherwise implies a contradictory chain of inequalities.
extbf{Second Approach}
- 2 points for proving that not all cells in a row/column and no transversal can satisfy the problem condition
- 2 points for proving that there exists a choice of \(a\) columns and \(b\) rows such that \(a + b = n + 1\)
- 2 points for correctly bounding the number of sapphire cells
|
USEMO-2021-2
|
https://web.evanchen.cc/exams/report-usemo-2021.pdf
|
Find all integers \( n \geq 1 \) such that \( 2^n - 1 \) has exactly \( n \) positive integer divisors.
|
The valid \( n \) are \( 1, 2, 4, 6, 8, 16, 32 \). They can be verified to work through inspection, using the well known fact that the Fermat prime \( F_i = 2^{2^i} + 1 \) is indeed prime for \( i = 0, 1, \ldots, 4 \) (but not prime when \( i = 5 \)).
We turn to the proof that these are the only valid values of \( n \). In both solutions that follow, \( d(n) \) is the divisor counting function.
Solution 1
Let \( d \) be the divisor count function. Now suppose \( n \) works, and write \( n = 2^k m \) with \( m \) odd. Observe that
\[
2^n - 1 = (2^m - 1)(2^m + 1)(2^{2m} + 1) \cdots (2^{2^{k-1}m} + 1),
\]
and all \( k+1 \) factors on the RHS are pairwise coprime. In particular,
\[
d(2^m - 1) d(2^m + 1) d(2^{2m} + 1) \cdots d(2^{2^{k-1}m} + 1) = 2^k m.
\]
Recall the following fact, which follows from MihÄilescuâs theorem.
**Lemma**: \( 2^r - 1 \) is a square if and only if \( r = 1 \), and \( 2^r + 1 \) is a square if and only if \( r = 3 \).
Now, if \( m \geq 5 \), then all \( k+1 \) factors on the LHS are even, a contradiction. Thus \( m \leq 3 \).
We deal with both cases.
If \( m = 1 \), then the inequalities
\[
\begin{aligned}
d(2^{2^0} - 1) &= 1 \
d(2^{2^0} + 1) &\geq 2 \
d(2^{2^1} + 1) &\geq 2 \
&\vdots \
d(2^{2^{k-1}} + 1) &\geq 2
\end{aligned}
\]
mean that it is necessary and sufficient for all of \( 2^{2^0} + 1, 2^{2^1} + 1, \ldots, 2^{2^{k-1}} + 1 \) to be prime. As mentioned at the start of the problem, this happens if and only if \( k \leq 5 \), giving the answers \( n \in \{1, 2, 4, 8, 16, 32\} \).
If \( m = 3 \), then the inequalities
\[
\begin{aligned}
d(2^{3 \cdot 2^0} - 1) &= 2 \
d(2^{3 \cdot 2^0} + 1) &= 3 \
d(2^{3 \cdot 2^1} + 1) &\geq 4 \
&\vdots \
d(2^{3 \cdot 2^{k-1}} + 1) &\geq 4
\end{aligned}
\]
mean that \( k \geq 2 \) does not lead to a solution. Thus \( k \leq 1 \), and the only valid possibility turns out to be \( n = 6 \).
Consolidating both cases, we obtain the claimed answer \( n \in \{1, 2, 4, 6, 8, 16, 32\} \).
Solution 2
Assume \( n \geq 7 \), and let \( n = \prod p_i^{e_i} \) be the prime factorization with \( e_i > 0 \) for each \( i \).
Define the numbers
\[
\begin{aligned}
T_1 &= 2^{p_1^{e_1}} - 1 \
T_2 &= 2^{p_2^{e_2}} - 1 \
&\vdots \
T_m &= 2^{p_m^{e_m}} - 1.
\end{aligned}
\]
We are going to use two facts about \( T_i \).
**Claim**: The \( T_i \) are pairwise relatively prime and
\[
\prod_{i=1}^m T_i \mid 2^n - 1.
\]
*Proof.* Each \( T_i \) divides \( 2^n - 1 \), and the relatively prime part follows from the identity \( \gcd(2^x - 1, 2^y - 1) = 2^{\gcd(x,y)} - 1 \).
**Claim**: The number \( T_i \) has at least \( e_i \) distinct prime factors.
*Proof.* This follows from Zsigmondyâs theorem: each successive quotient \( (2^{p^{k+1}} - 1)/(2^{p^k} - 1) \) has a new prime factor.
**Claim (Main claim)**: Assume \( n \) satisfies the problem conditions. Then both the previous claims are sharp in the following sense: each \( T_i \) has *exactly* \( e_i \) distinct prime divisors, and
\[
\left\{ \text{primes dividing } \prod_{i=1}^m T_i \right\} = \left\{ \text{primes dividing } 2^n - 1 \right\}.
\]
*Proof.* Rather than try to give a size contradiction directly from here, the idea is to define an ancillary function
\[
s(x) = \sum_{p \text{ prime}} \nu_p(x)
\]
which computes the sum of the exponents in the prime factorization. For example
\[
s(n) = e_1 + e_2 + \cdots + e_m.
\]
On the other hand, using the earlier claim, we get
\[
s(d(2^n - 1)) \geq s\left( d\left(\prod T_i\right) \right) \geq e_1 + e_2 + \cdots + e_m = s(n).
\]
But we were told that \( d(2^n - 1) = n \); hence equality holds in all our estimates, as needed.
At this point, we may conclude directly that \( m = 1 \) in any solution; indeed if \( m \geq 2 \) and \( n \geq 7 \), Zsigmondyâs theorem promises a primitive prime divisor of \( 2^n - 1 \) not dividing any of the \( T_i \).
Now suppose \( n = p^e \), and \( d(2^{p^e} - 1) = n = p^e \). Since \( 2^{p^e} - 1 \) has exactly \( e \) distinct prime divisors, this can only happen if in fact
\[
2^{p^e} - 1 = q_1^{p-1} q_2^{p-1} \cdots q_e^{p-1}
\]
for some distinct primes \( q_1, q_2, \ldots, q_e \). This is impossible modulo 4 unless \( p = 2 \).
So we are left with just the case \( n = 2^e \), and need to prove \( e \leq 5 \). The proof consists of simply remarking that \( 2^{2^5} + 1 \) is known to not be prime, and hence for \( e \geq 6 \) the number \( 2^{2^e} - 1 \) always has at least \( e+1 \) distinct prime factors.
|
In this rubric, none of the items are additive: neither the positive items nor the deductions. Hence an incomplete solution receives the largest positive item, while a complete solution receives 7 minus the largest deduction. Deductions do not apply to solutions scoring at most 2 points.
Common items for both solutions
- 0 points for correct solution set.
- 0 points for proving that special cases of \( n \) (odd \( n \), or prime power \( n \)) donât work.
- -1 points for not mentioning anywhere that all \( n \) in the solution set work.
- -1 points for a solution which claims that all powers of 2 work (but resolves the other cases correctly). (Stating that itâs well-known \( F_i \) is prime for \( i = 1, \ldots, 4 \) but not 5, where \( F_i \) are the Fermat primes, counts as a correct proof.)
- -1 points for a solution which has an incorrect solution set but is otherwise correct (unless the only error is missing \( n = 1 \), in which case there is no deduction).
First official solution
- 0 points for just writing down \( 2^n - 1 = (2^m - 1)(2^m + 1)(2^{2m} + 1) \cdots \).
- 2 points for proving that one of \( 2^m - 1, 2^m + 1, 2^{2m} + 1, \ldots \) is a square.
- 3 points for proving that \( m = 1 \) or 3.
- 7 points for a complete solution.
- -1 points for an incorrect proof, or statement without proof, that \( 2^r - 1 \) is only a square when \( r = 1 \), and/or that \( 2^r + 1 \) is only a square when \( r = 3 \), if the solution is otherwise correct. (Citing Catalan/MihÄilescu counts as a correct proof.)
Second official solution
- 0 points for just writing down a result of Zsigmondy (that \( 2^n - 1 \) has at least \( e_1 + \cdots + e_m \), or \( d(n) - 2 \), distinct prime factors).
- 2 points for proving that equality holds in the estimate \( s(d(2^n - 1)) \geq \cdots \) or something similar.
- 3 points for proving \( n \) is 6 or a prime power.
- 7 points for a complete solution.
|
USEMO-2021-3
|
https://web.evanchen.cc/exams/report-usemo-2021.pdf
|
Let \( A_1C_2B_1A_2C_1B_2 \) be an equilateral hexagon. Let \( O_1 \) and \( H_1 \) denote the circumcenter and orthocenter of \( riangle A_1B_1C_1 \), and let \( O_2 \) and \( H_2 \) denote the circumcenter and orthocenter of \( riangle A_2B_2C_2 \). Suppose that \( O_1
e O_2 \) and \( H_1
e H_2 \). Prove that the lines \( O_1O_2 \) and \( H_1H_2 \) are either parallel or coincide.
|
Let \( riangle X_1Y_1Z_1 \) and \( riangle X_2Y_2Z_2 \) be the medial triangles of \( riangle A_1B_1C_1 \) and \( riangle A_2B_2C_2 \). The first simple observation is as follows.
\begin{quote}
\textbf{Claim} â \( Y_1, Z_1, Y_2, Z_2 \) are concyclic.
\end{quote}
\textit{Proof.} The distance from each of \( Y_1, Z_1, Y_2, Z_2 \) to the midpoint of \( \overline{A_1A_2} \) is half the side length of the hexagon. \(\square\)
Hence by radical axis argument, we obtain that \( X_1X_2, Y_1Y_2, Z_1Z_2 \) are concurrent, except possibly when all six points lie on a circle. In this case, \( \triangle A_1B_1C_1 \) and \( \triangle A_2B_2C_2 \) share the same nine-point center, so clearly \( O_1O_2 \parallel H_1H_2 \). So we will assume going forward that \( (X_1Y_1Z_1) \) and \( (X_2Y_2Z_2) \) are distinct circles.
\begin{quote}
\textbf{Claim (Perspectivity)} â The two triangles \( \triangle X_1Y_1Z_1 \) and \( \triangle X_2Y_2Z_2 \) are perspective at some point \( K \).
\end{quote}
\textit{Proof.} As mentioned above, \( X_1X_2, Y_1Y_2, Z_1Z_2 \) are concurrent. \(\square\)
Let \( N_1 \) and \( G_1 \) be the circumcenter and centroid of \( \triangle X_1Y_1Z_1 \); define \( N_2 \) and \( G_2 \) similarly.
\begin{quote}
\textbf{Claim (Orthology)} â Triangles \( \triangle X_1Y_1Z_1 \) and \( \triangle X_2Y_2Z_2 \) are orthologic. In fact, the orthology center \( S_1 \) is the image of \( O_2 \) under a homothety centered at \( G_1 \) with ratio \( -\tfrac{1}{2} \).
\end{quote}
\textit{Proof.} Since the mentioned homothety takes \( \overrightarrow{A_1O_2} \to \overrightarrow{X_1S_1} \), so
\[
Y_2Z_2 \parallel B_2C_2 \perp A_1O_2 \parallel X_1S_1
\]
as desired. \(\square\)
We have obtained that \( \triangle X_1Y_1Z_1 \) and \( \triangle X_2Y_2Z_2 \) are both orthologic (with centers \( S_1 \) and \( S_2 \)) and perspective (through \( K \)). Hence it follows by \textbf{Sondatâs theorem} that \( S_1, S_2, K \) lie on a line perpendicular to the perspectrix.
To finish, we follow up with two more claims:
\begin{quote}
\textbf{Claim (Perspectrix is radical axis)} â The perspectrix of the two triangles is exactly the radical axis of their circumcircles, hence perpendicular to \( N_1N_2 \).
\end{quote}
\textit{Proof.} This follows from the earlier observation that \( Y_1, Y_2, Z_1, Z_2 \) was cyclic, etc. \(\square\)
\begin{quote}
\textbf{Claim (Degenerate parallelogram)} â \( N_1S_1N_2S_2 \) is a (possibly degenerate) parallelogram.
\end{quote}
\textit{Proof.} Because
\[
\overrightarrow{S_1N_2} = \overrightarrow{O_2G_2} = \tfrac{3}{2}\overrightarrow{G_1G_2} = \overrightarrow{O_1G_1} = \overrightarrow{N_1S_2}.
\]
In this way we can conclude that \( \overline{N_1N_2} \parallel \overline{S_1S_2} \) through the former claim, but they have the same midpoint by the latter claim, so ultimately all \( N_i \) and \( S_i \) are collinear.
Finally, note that
\[
\overline{N_1N_2} \parallel \overline{N_1S_1} \parallel \overline{G_1O_1} \parallel \overline{O_1O_2}.
\]
It easily follows that \( \overline{O_1O_2} \parallel \overline{H_1H_2} \), as wanted. \(\square\)
\textbf{Remark.} An amusing corollary of the above solution is the following:
\textit{Assuming \( A_1C_2B_1A_2C_1B_2 \) is not self-intersecting, the midpoints of \( A_1A_2, B_1B_2, C_1C_2 \) cannot be collinear} (unless two of them coincide).
To see this, let \( M_A, M_B, M_C \) be said midpoints. If they are different and lie on line \( \ell \), then \( M_BX_1M_CX_2 \) is a rhombus with side length \( \tfrac{1}{2}s \), so \( X_1 \) and \( X_2 \) are reflections in \( \ell \).
Similarly, \( \triangle X_1Y_1Z_1 \) and \( \triangle X_2Y_2Z_2 \) are reflections in \( \ell \), so \( \triangle A_1B_1C_1 \) and \( \triangle A_2B_2C_2 \) are as well. This is not possible if \( A_1C_2B_1A_2C_1B_2 \) is not self-intersecting, because some side will intersect \( \ell \): then its opposite side will intersect this side at the intersection point.
|
Because the number of solutions with any substantial progress was low, there was no official rubric written for this problem. Instead, the graders first identified all papers that were plausible candidates for partial marks, if any. Then they discussed each individual paper case by case.
|
USEMO-2021-4
|
https://web.evanchen.cc/exams/report-usemo-2021.pdf
|
Let \( ABC \) be a triangle with circumcircle \( \omega \), and let \( X \) be the reflection of \( A \) in \( B \). Line \( CX \) meets \( \omega \) again at \( D \). Lines \( BD \) and \( AC \) meet at \( E \), and lines \( AD \) and \( BC \) meet at \( F \). Let \( M \) and \( N \) denote the midpoints of \( AB \) and \( AC \). Can line \( EF \) share a point with the circumcircle of triangle \( AMN \)?
|
The answer is no, they never intersect.
**Solution 1 (Classical solution, by author)**
Let \( P \) denote the midpoint of \( \overline{AD} \), which
- lies on \( \overline{BN} \), since \( BN \parallel CX \); and
- lies on \( (AMN) \), since itâs homothetic to \( (ABC) \) through \( A \) with factor \( \tfrac{1}{2} \).
Now, note that
\[
\angle FBP = \angle CBN = \angle BCD = \angle BAD = \angle BAF \implies FB^2 = FP \cdot FA
\]
\[
\angle EBN = \angle EDC = \angle BDC = \angle BAC = \angle BAE \implies EB^2 = EN \cdot EA
\]
This means that line \( EF \) is the radical axis of the circle centered at \( B \) with radius zero, and the circumcircle of triangle \( AMN \). Since \( B \) obviously lies outside \( (AMN) \), the disjointness conclusion follows.
**Solution 2 (Projective solution, by Ankit Bisain)**
In this approach we are still going to prove that \( \overline{EF} \) is the radical axis of \( (AMN) \) and the circle of radius zero at \( B \), but we are not going to use the point \( P \), or even points \( E \) and \( F \).
Instead, let \( Y = \overline{EF} \cap \overline{AB} \), which by Brokardâs theorem on \( ABDC \) satisfies \( (AB;XY) = -1 \). Since \( XB = XA \), it follows that \( AY : YB = 2 \). From here it is straightforward to verify that
\[
YB^2 = \tfrac{1}{9} AB^2 = YM \cdot YA
\]
Thus \( Y \) lies on the radical axis.
Finally, by Brokardâs theorem again, if \( O \) is the center of \( \omega \) then \( \overline{OX} \perp \overline{EF} \). Taking a homothety with scale factor 2 at \( A \), it follows that the line through \( B \) and the center of \( (AMN) \) is perpendicular to \( \overline{EF} \).
Since \( \overline{EF} \) contains \( Y \), it now follows that \( \overline{EF} \) is the radical axis, as claimed.
**Solution 3 (Solution with inversion, projective, and Cartesian coordinates, by Ankan Bhattacharya)**
In what follows, let \( O \) be the center of \( \omega \). Note that Brokardâs theorem gives that \( \overline{EF} \) is the polar of \( X \).
Note that since none of \( E, F, X \) are points at infinity, \( O \) is different from all three. We consider inversion in \( \omega \) to eliminate the polar:
- The circumcircle of \( \triangle AMN \), i.e. the circle with diameter \( \overline{AO} \), is sent to the line \( \ell \) tangent to \( \omega \) at \( A \).
- The line \( EF \), as the polar of \( X \), is sent to the circle with diameter \( \overline{OX} \). (It is indeed a circle, because \( O \) does not lie on line \( EF \).)
Thus, if the posed question is true, then we see that \( \ell \) intersects \( (OX) \). We claim this is impossible.
Establish Cartesian coordinates with \( A = (0,0) \) and \( O = (2,0) \), so \( \ell \) is the y-axis. Let \( T \) be the center of \( (OX) \): the midpoint of \( \overline{OX} \). Observe:
- \( B \) lies on the circle with center \( (2,0) \) and radius 2.
- \( X \) lies on the circle with center \( (4,0) \) and radius 4.
- \( T \) lies on the circle with center \( (3,0) \) and radius 2.
Thus, let the coordinates of \( T \) be \( (x,y) \), with \( (x-3)^2 + y^2 = 4 \). The intersection of \( \ell \) and \( (OX) \) being nonempty is equivalent to
\[
d(T, \ell)^2 \le OT^2 \
\iff x^2 \le (x-2)^2 + y^2 \
\iff x^2 \le (x-2)^2 + [4 - (x-3)^2] \
\iff (x-1)^2 \le 0,
\]
or \( x = 1 \) (which forces \( y = 0 \)); i.e. \( T = (1,0) \). However, this forces
\[
B = (0,0) = A,
\]
which is not permitted. Thus, line \( \ell \) cannot share a point with \( (OX) \), and so line \( EF \) cannot share a point with \( (AMN) \).
|
The following things might happen:
(a) It is stated/conjectured that answer is NO.
(b) It is mentioned that \( \overline{EF} \) is the polar of \( X \) wrt \( \omega \).
(c) Point \( K = \overline{EF} \cap \overline{AB} \) is introduced.
(d) It is mentioned that \( rac{AK}{KB} = 2 \).
(e) Center \( O \) of \( \omega \) is defined.
(f) As we only need to show distance between midpoint of segment \( AO \) and line \( EF \) is \( R/2 \) (where \( R \) is the radius of \( \omega \)), so it is mentioned that it suffices to show
\[
d(A, \overline{EF}) + d(O, \overline{EF}) > R
\]
(g) Now in the above step, we require that \( A, O \) lie on the same side of line \( EF \). So some students might not mention that.
(h) Proving (1).
Now we can give the following marks:
(i) 0 marks for only (a), (b).
(ii) 1 marks for (c), (d).
(iii) 2 marks for mentioning (1) without doing the mistake in (g) and also proving \( d(A, \overline{EF}) = rac{d(X, \overline{EF})}{2} \).
(iv) 1 marks if mistake in (g) is also done in (f).
(v) 6 marks If the solution is correct till end but mistake in (g) is done.
(vi) 7 marks for a perfect solution.
|
USEMO-2021-5
|
https://web.evanchen.cc/exams/report-usemo-2021.pdf
|
Given a polynomial \( p(x) \) with real coefficients, we denote by \( S(p) \) the sum of the squares of its coefficients. For example, \( S(20x + 21) = 20^2 + 21^2 = 841 \).
Prove that if \( f(x), g(x), \) and \( h(x) \) are polynomials with real coefficients satisfying the identity \( f(x) \cdot g(x) = h(x)^2 \), then
\[
S(f) \cdot S(g) \ge S(h)^2.
\]
|
Solution 1
Claim â Let \( p \) be a polynomial with real coefficients, and \( n > \deg p \) an integer. Then
\[
S(p) = \frac{1}{n} \sum_{k=0}^{n-1} \left|p\left(e^{2\pi i k/n}\right)\right|^2.
\]
Proof. Note that
\[
\left|p\left(e^{2\pi i k/n}\right)\right|^2 = p\left(e^{2\pi i k/n}\right) \cdot p\left(e^{-2\pi i k/n}\right)
\]
so if we define \( q(x) = p(x)p(1/x) \), the right-hand side is the sum of \( q \) across the \( n \)th roots of unity.
Applying a roots of unity filter, the right-hand side is the constant coefficient of \( q(x) \). But that constant coefficient is exactly equal to \( S(p) \).
To solve the problem, choose \( n > \max\{\deg f, \deg g, \deg h\} \), set \( \omega = e^{2\pi i/n} \), and apply the key claim to all three to get that the desired inequality is equivalent to
\[
\left[ \frac{1}{n} \sum \left|f(\omega^k)\right|^2 \right] \cdot \left[ \frac{1}{n} \sum \left|g(\omega^k)\right|^2 \right] \ge \left[ \frac{1}{n} \sum \left|h(\omega^k)\right|^2 \right]^2
\]
\[\iff \left[ \sum \left|f(\omega^k)\right|^2 \right] \cdot \left[ \sum \left|g(\omega^k)\right|^2 \right] \ge \left[ \sum \left|f(\omega^k)\right| \cdot \left|g(\omega^k)\right| \right]^2.\]
This is just Cauchy-Schwarz, so we are done.
Remark (Continuous version of above solution). To avoid the arbitrary choice of parameter \( n \), one can make the same argument to show that for any \( p \in \mathbb{R}[x] \),
\[
S(p) = \frac{1}{2\pi} \int_0^{2\pi} \left|p(e^{ix})\right|^2 dx.
\]
Using Cauchyâs inequality for integrals, we obtain a continuous version of the above solution. However, this is technically out of scope for high-school olympiad, despite the fact it is really just the limit as \( n \to \infty \) of the above solution.
|
Roots of unity solution
The following items apply but are not additive:
- 0 points for proving the inequality for special cases. For example when \( f \) is a quadratic polynomial, cubic polynomial etc.
- 0 points for expressing the polynomials as \( f(x) = p(x)^2r(x) \), \( g(x) = q(x)^2r(x) \) and \( h(x) = p(x)q(x)r(x) \).
- 1 point for proving that \( S(p) \) is equal to the independent coefficient of \( p(x)p(1/x) \)
- 5 points for using a roots of unity filter to characterize \( S(p) \) as
\[
S(p) = \frac{1}{n} \sum_{i=0}^{n-1} |p(e^{2\pi i k/n})|^2 \quad \text{or} \quad S(p) = \frac{1}{2\pi} \int_0^{2\pi} |p(e^{ix})|^2 dx
\]
- 7 points for a complete solution
As always there will be a 1 point deduction to essentially complete solutions with minor errors that could be easily be fixed such as:
- not picking a big enough \( n \) when proving or applying the discrete characterisation of \( S(p) \).
Coefficient manipulation solution
The following items apply but are not additive:
- 0 points for proving the inequality for special cases. For example when \( f \) is a quadratic polynomial, cubic polynomial etc.
- 0 points for expressing the polynomials as \( f(x) = p(x)^2r(x) \), \( g(x) = q(x)^2r(x) \) and \( h(x) = p(x)q(x)r(x) \).
- 1 point for proving that \( S(p) \) is equal to the independent coefficient of \( p(x)p(1/x) \)
- 5 points for proving the identities
\[
S(r(x)p(x)^2) = \sum_k \left( r(x)p(x)p\left(\frac{1}{x}\right)[x^k] \right)^2
\]
\[
S(r(x)q(x)^2) = \sum_k \left( r\left(\frac{1}{x}\right) q(x)q\left(\frac{1}{x}\right) [x^{-k}] \right)^2
\]
- 7 points for a complete solution
|
USEMO-2021-6
|
https://web.evanchen.cc/exams/report-usemo-2021.pdf
|
A \emph{bagel} is a loop of \( 2a + 2b + 4 \) unit squares which can be obtained by cutting a concentric \( a imes b \) hole out of an \( (a+2) imes (b+2) \) rectangle, for some positive integers \( a \) and \( b \). (The side of length \( a \) of the hole is parallel to the side of length \( a+2 \) of the rectangle.)
Consider an infinite grid of unit square cells. For each even integer \( n \geq 8 \), a \emph{bakery of order \( n \)} is a finite set of cells \( S \) such that, for every \( n \)-cell bagel \( B \) in the grid, there exists a congruent copy of \( B \) all of whose cells are in \( S \). (The copy can be translated and rotated.) We denote by \( f(n) \) the smallest possible number of cells in a bakery of order \( n \).
Find a real number \( lpha \) such that, for all sufficiently large even integers \( n \geq 8 \), we have
\[
rac{1}{100} < rac{f(n)}{n^lpha} < 100.
\]
|
The answer is \( lpha = 3/2 \).
In what follows, â\( Y \) is about \( X \)â means that \( Y = [1+o(1)]X \). Equivalently, \( \lim_{n o \infty} Y/X = 1 \).
1. Intuitively, both of these say that \( X \) and \( Y \) become closer and closer together as \( n \) grows. This is fine for the problem since only sufficiently large \( n \) are involved.
¶ Bound. First we prove that every bakery \( S \) of order \( n \) contains at least about \( n^{3/2}/8 \) cells.
We say that a bagel is \emph{horizontal} or \emph{vertical} depending on the orientation of its pair of longer sides. (A square bagel is both.) For each \( a < b \) with \( 2a + 2b + 4 = n \), take one bagel in \( S \) whose hole is of size either \( a imes b \) or \( b imes a \). Without loss of generality, at least about \( n/8 \) of our bagels are horizontal.
Say that there are a total of \( k \) rows which contain a longer side of at least one of our horizontal bagels. Note that the shorter side length of a horizontal bagel depends only on the distance between the rows of its longer sides. Since the shorter side lengths of all of our bagels are pairwise distinct, we obtain that \( inom{k}{2} \) is at least about \( n/8 \). Consequently, \( k \) is at least about \( \sqrt{n}/2 \).
On the other hand, each such row contains at least about \( n/4 \) cells in \( S \). Therefore, \( |S| \) is at least about \( n^{3/2}/8 \), as needed.
¶ Construction. To complete the solution, we construct a bakery \( S \) of order \( n \) with at most about \( \sqrt{2} \cdot n^{3/2} \) cells. Define
\[
\ell = \left\lceil \sqrt{n/2}
ight
ceil \quad ext{and} \quad D = \{-\ell^2, -(\ell-1)\ell, \ldots, -3\ell, -2\ell, -\ell, 0, 1, 2, \ldots, \ell\}.
\]
Then \( |D| \) is about \( \sqrt{2n} \).
We refer to the set \( D \) as a \emph{ruler} in the sense that for any \( 1 \leq m < n/2 \), there are \( x_1 \) and \( x_2 \) in \( D \) with \( x_2 - x_1 = m \). Indeed, one lets \( x_2 \) be the remainder when \( m \) is divided by \( \ell \), so that \( x_1 = x_2 - m \leq 0 \) is a multiple of \( \ell \).
Now, if we let \( T = \{-\ell^2, -\ell^2 + 1, \ldots, \ell\} \) then we may define
\[
S = (D imes T) \cup (T imes D).
\]
An illustration below is given for \( \ell = 5 \).
Note that \( |S| \) is at most about \( n|D| \), that is, at most about \( \sqrt{2} \cdot n^{3/2} \).
extbf{Claim â} The set \( S \) is a bakery of order \( n \).
extit{Proof.} Let \( a \) and \( b \) be any positive integers with \( 2a + 2b + 4 = n \). By the choice of \( D \), there are \( x_1 \) and \( x_2 \) in \( D \) such that \( x_2 - x_1 = a + 1 \), as well as \( y_1 \) and \( y_2 \) in \( D \) such that \( y_2 - y_1 = b + 1 \). Then the bagel with opposite corner cells \( (x_1, y_1) \) and \( (x_2, y_2) \) has a hole with side lengths \( a \) and \( b \) and all of its cells are in \( S \), as needed. \( \square \)
extbf{Remark.} Let us call a ruler \emph{sparse} when a lot of its marks are missing but we can still measure out each one of the distances \( 1, 2, \ldots, N \). Then for the set \( D \) in the solution essentially we need a sparse ruler with about \( c\sqrt{N} \) marks, for some reasonably small positive real constant \( c \). The construction above is simple but also far from optimal. Other constructions are known which are more complicated but yield smaller values of \( c \). See, for example, Ed Pegg Jr, \emph{Hitting All The Marks}.
|
No points are awarded for the answer \( lpha = 3/2 \) alone. However, the following items are possible:
- A complete construction is worth extbf{2 points}. A student can earn extbf{1 point} of this item for stating the answer \( lpha = 3/2 \) roughly describing the idea of the construction (that is, to find a ârulerâ \( D \), and hope that the ruler has about \( \Theta(\sqrt{n}) \) numbers). This can be given even if the student has no idea how to actually find such a ruler.
- Meanwhile, proving the lower bound \( lpha \geq 3/2 \) is worth extbf{2 points}.
These two items are additive, meaning \( 1 + 2 = 3 \) while \( 2 + 2 = 7 \).
|
USEMO-2022-1
|
https://web.evanchen.cc/exams/report-usemo-2022.pdf
|
A \emph{stick} is defined as a \(1 imes k\) or \(k imes 1\) rectangle for any integer \(k \geq 1\). We wish to partition the cells of a \(2022 imes 2022\) chessboard into \(m\) non-overlapping sticks, such that any two of these \(m\) sticks share at most one unit of perimeter. Determine the smallest \(m\) for which this is possible.
|
In general, with 2022 replaced by \(n\), we will prove the answer is
\[
m = egin{cases}
1 & ext{if } n = 1 \
3 & ext{if } n = 2 \
rac{1}{2}(n^2 - 2n + 7) & ext{if } n \geq 3 ext{ and } n ext{ is odd} \
rac{1}{2}(n^2 - 2n + 8) & ext{if } n \geq 3 ext{ and } n ext{ is even}
\end{cases}
\]
with the following construction. For \(n = 2022\) this gives 2042224 as the answer.
The optimality for \(n \in \{1,2\}\) is easy to check, so assume \(n \geq 3\).
The main idea is to view the problem as taking an \(n imes n\) grid of squares and deleting some of the edges until the resulting figure is a set of sticks. Since the number of sticks is equal to \(n^2\) minus the number of removed edges in the square, it suffices to maximize the number of removed edges. However:
extbf{Claim} â No two of the deleted edges may share an endpoint.
extit{Proof.} Obvious. \(\square\)
As a consequence of this claim:
- At most \( rac{1}{2}(n - 3)^2\) edges can be removed within the central \((n - 3) imes (n - 3)\) grid of lattice points.
- Additionally, at most \(4(n - 2)\) of the outer edges can be removed since there are \(4(n - 2)\) lattice points within one unit of the boundary.
Hence, the minimum number of removed edges is at least
\[
n^2 - \left(\left\lfloor rac{1}{2}(n - 3)^2
ight
floor + 4(n - 2)
ight)
\]
which equals the claimed minimum.
extbf{Remark (Torus variant).} One can ask the same problem on an \(n imes n\) torus. The answer is \( rac{1}{2}n^2\) for even \(n\) and \( rac{1}{2}(n^2 - 1)\) for odd \(n\); the proof is analogous to the one above, but without the outer edge consideration.
|
For solutions which are not complete, the following items are available and are additive:
(i) \(+1\) point is awarded for stating the correct answer which is 2,042,224 (or the answer for general \(n\)), even with no justification.
(ii) \(+1\) point is awarded for giving a working construction.
(iii) \(+1\) point for the observation that there is at most one deleted edge per vertex.
For weaker bounds of \(m\) (if the value of \(m\) is not correct), the following items are available, which are not additive with (i)-(iii) or each other:
(iv) \(1\) point for constructions which use at most
\[
rac{1}{2}(n^2 - n) + 2 = 2043233
\]
sticks; an example of such a construction is shown below for \(n = 8\).
Note that, for odd \(n\), this construction is actually optimal, so the student would have gotten 2 points in an odd year!
(v) \(0\) points for constructions which use more than 2043233 sticks.
(vi) \(1\) point for proving that \(m \geq cn^2\) is extit{necessary} to fulfill the condition (i.e. a bound, not a construction), for any positive constant \(c \geq 1/3\).
(vii) \(2\) points for proving that \(m \geq rac{1}{2}n^2 - cn\) is extit{necessary} to fulfill the condition (i.e. a bound, not a construction), for any positive constant \(c \geq 1\).
And obviously, \(7\) points for a perfectly working solution. The following deductions can apply to a correct solution; they are additive.
(viii) \(-1\) point if whole solution is correct but there are calculation/algebraic errors in answer extraction (e.g. calculating for \(n = 2022\), algebraic errors)
(ix) \(-1\) point if the construction given (if any) is wrong or missing.
No deduction for not stating the answer for \(n = 2022\) explicitly (if they did the problem by considering a general \(n\) and there are no other errors).
|
USEMO-2022-2
|
https://web.evanchen.cc/exams/report-usemo-2022.pdf
|
A function \( \psi : \mathbb{Z} o \mathbb{Z} \) is said to be \emph{zero-requiem} if for any positive integer \( n \) and any integers \( a_1, \ldots, a_n \) (not necessarily distinct), the sums \( a_1 + a_2 + \cdots + a_n \) and \( \psi(a_1) + \psi(a_2) + \cdots + \psi(a_n) \) are not both zero.
Let \( f \) and \( g \) be two zero-requiem functions for which \( f \circ g \) and \( g \circ f \) are both the identity function (that is, \( f \) and \( g \) are mutually inverse bijections). Given that \( f + g \) is \emph{not} a zero-requiem function, prove that \( f \circ f \) and \( g \circ g \) are both zero-requiem.
|
Solution 1 (rephrased from several contestants).
Assume for contradiction that neither \( f \circ f \) nor \( g \circ g \) is zero-requiem, meaning that there exist \( c_1, \ldots, c_m \) and \( w_1, \ldots, w_n \) such that
\[
0 = \sum_{i=1}^m c_i = \sum_{i=1}^m f(c_i) + \sum_{i=1}^m g(c_i)
\]
\[
0 = \sum_{i=1}^n w_i = \sum_{i=1}^n g(g(w_i)).
\]
Set \( x_i = g(w_i) \) and discard \( w_i \); then we can rewrite the two equations as
\[
0 = \sum_{i=1}^m c_i = \sum_{i=1}^m f(c_i) + \sum_{i=1}^m g(c_i) \quad \text{\small (\( \neq 0 \) as \( f \) is ZR, \( \neq 0 \) as \( g \) is ZR)}
\]
\[
0 = \sum_{i=1}^n f(x_i) = \sum_{i=1}^n g(x_i).
\]
Since \( f \) and \( g \) were given to be zero-requiem, the sums \( \sum_{i=1}^m f(c_i) \), \( \sum_{i=1}^m g(c_i) \), and \( \sum_{i=1}^n x_i \) are all nonzero. So let’s say WLOG that \( \sum_{i=1}^m g(c_i) \) and \( \sum_{i=1}^m x_i \) have opposite sign. That means we have integers \( p, q > 0 \) such that
\[
p \sum_{i=1}^m c_i + q \sum_{i=1}^n x_i = 0.
\]
But
\[
p \sum_{i=1}^m c_i + q \sum_{i=1}^n f(x_i) = p \cdot 0 + q \cdot 0 = 0.
\]
This is a contradiction to \( f \) being zero-requiem if we feed in the sequence constructed by taking \( (c_1, \ldots, c_m) \) each \( p \) times and \( (x_1, \ldots, x_n) \) each \( q \) times.
Solution 2 (author’s).
We say a function \( f : \mathbb{Z} \to \mathbb{Z} \) is
- \textbf{Nunnally} if \( a_1 + \cdots + a_n \ge 0 \) implies \( f(a_1) + \cdots + f(a_n) > 0 \);
- \textbf{Marianne} if \( a_1 + \cdots + a_n \ge 0 \) implies \( f(a_1) + \cdots + f(a_n) < 0 \);
- \textbf{Jeremiah} if \( a_1 + \cdots + a_n \le 0 \) implies \( f(a_1) + \cdots + f(a_n) > 0 \);
- \textbf{Charles} if \( a_1 + \cdots + a_n \le 0 \) implies \( f(a_1) + \cdots + f(a_n) < 0 \);
for any finite sequence \( a_1, \ldots, a_n \) of integers.
\textbf{Claim.} A function is zero-requiem if and only if it is at least one of these four categories.
\textit{Proof.} Using the condition on the sequence \( (0,0) \) we see that \( f(0) \ne 0 \). Thus there are two cases, \( f(0) > 0 \) and \( f(0) < 0 \). These cases are essentially the same, so we start with \( f(0) > 0 \).
When \( f(0) > 0 \), there are three sub-cases:
- Suppose there exists a positive integer \( x \) so that \( f(x) \le 0 \), say \( f(x) = -y \). We claim that \( f \) is a Jeremiah. Indeed, if not, then there exist \( a_1, \ldots, a_n \) with non-positive sum \( -p \), so that \( f(a_1) + \cdots + f(a_n) = -S \) is non-positive. Consider the following sequence:
\[
\underbrace{a_1, \ldots, a_n, \ldots, a_1, \ldots, a_n}_{x f(0) \text{ blocks}}, \underbrace{x, \ldots, x}_{pf(0) \text{ x's}}, \underbrace{0, \ldots, 0}_{xS + py \text{ 0's}}.
\]
Clearly this has sum \( x f(0) \cdot (-p) + pf(0) \cdot x + 0 = 0 \). The sum of the sequence of the \( f \)-values of these is
\[
x f(0) \cdot (f(a_1) + \cdots + f(a_n)) + pf(0) \cdot f(x) + (xS + py) f(0) \
= x f(0)(-S) + pf(0)(-y) + (xS + py) f(0) = 0.
\]
This is a contradiction, proving our claim.
- Suppose there exists a negative integer \( x \) so that \( f(x) \le 0 \). Now the function \( f_1 \) defined by \( f_1(x) = f(-x) \) satisfies the hypotheses of the previous case, so \( f_1 \) is a Jeremiah. Then clearly \( f \) is a Nunnally.
- Otherwise \( f(x) > 0 \) for all \( x \), so \( f \) is trivially a Nunnally.
Now say \( f(0) < 0 \). Then \( f_2 = -f \) is also a zero-requiem and satisfies \( f_2(0) > 0 \), so it’s either a Jeremiah or a Nunnally, whence clearly \( f \) is either a Charles or Marianne. This proves our claim. \( \square \)
\textbf{Claim.} The following four statements are true.
- If \( f \) is a Nunnally or a Jeremiah, \( f^{-1} \) cannot be Nunnally or a Marianne.
- If \( f \) is a Marianne or a Charles, \( f^{-1} \) cannot be a Jeremiah or a Charles.
- If \( f^{-1} \) is a Nunnally or a Jeremiah, \( f \) cannot be Nunnally or a Marianne.
- If \( f^{-1} \) is a Marianne or a Charles, \( f \) cannot be a Jeremiah or a Charles.
\textit{Proof.} For the first statement, note that
\[
a_1 + \cdots + a_n = 0 \implies f(a_1) + \cdots + f(a_n) > 0 \
\implies f^{-1}(f(a_1)) + \cdots + f^{-1}(f(a_n)) \ne 0 \
\implies a_1 + \cdots + a_n \ne 0.
\]
The second statement is proven in the same way. The third and fourth statements follow by swapping the roles of \( f \) and \( f^{-1} \) in the first two statements. \( \square \)
Back to the main problem.
\textbf{Claim.} If \( f \) is a Nunnally or a Charles, then \( f \circ f \) is a zero-requiem.
\textit{Proof.} Indeed, in the first case, we have
\[
a_1 + \cdots + a_n = 0 \implies f(a_1) + \cdots + f(a_n) > 0 \
\implies f(f(a_1)) + \cdots + f(f(a_n)) > 0 \
\implies (f \circ f)(a_1) + \cdots (f \circ f)(a_n) \ne 0,
\]
proving the claim. The other case follows similarly.
The only remaining cases are when \( f \) and \( f^{-1} \) are both Mariannes or they are both Jeremiahs. In the first case,
\[
a_1 + \cdots + a_n = 0 \implies f(a_1) + \cdots + f(a_n) < 0 \text{ and } f^{-1}(a_1) + \cdots f^{-1}(a_n) < 0 \
\implies (f(a_1) + f^{-1}(a_1)) + \cdots + (f(a_n) + f^{-1}(a_n)) < 0 \
\implies (f + f^{-1})(a_1) + \cdots + (f + f^{-1})(a_n) \ne 0,
\]
which means \( f + f^{-1} \) is zero-requiem. A similar argument holds when \( f, f^{-1} \) are both Jeremiahs, so we are done. \( \square \)
|
(i) 4 points: Proving the first lemma in the first solution
(ii) 1 point: Conjecturing the first lemma in the first solution
(iii) 0 points: Neither \( \sum_{i=1}^k f(a_i) \) nor \( \sum_{i=1}^k g(a_i) \) is zero.
(iv) 1 point: For showing that for all \( (a_n)_n \) such \( a_1 + \cdots + a_n = 0 \), \( f(a_1) + \cdots + f(a_n) \) are either all positive or all negative.
(v) 2 points: For proving \( f(x) > 0 \) must hold for either all positive \( x \) or all negative \( x \)
(vi) 1 point: For considering \( f(a_1) + \cdots + f(a_k) = g(f \circ f(a_1)) + \cdots + g(f \circ f(a_k)) \) and \( f, g \) being opposite signs
(vii) 3 points**: Proof of existence of \( \sum_{i=1}^\ell b_i \) that has opposite sign with either \( \sum_{i=1}^k f(a_i) \) or \( \sum_{i=1}^k g(a_i) \) (Only for proofs by contradiction)
(viii) 5 points: Failing to finish in the case \( f(x) > 0 \) for \( x < 0 \), but otherwise complete
(ix) 5 points: Failing to finish in the case \( f(x) > 0 \) for \( x > 0 \), but otherwise complete
|
USEMO-2022-3
|
https://web.evanchen.cc/exams/report-usemo-2022.pdf
|
Point \( P \) lies in the interior of a triangle \( ABC \). Lines \( AP, BP, \) and \( CP \) meet the opposite sides of triangle \( ABC \) at points \( A', B', \) and \( C' \), respectively. Let \( P_A \) be the midpoint of the segment joining the incenters of triangles \( BPC' \) and \( CPB' \), and define points \( P_B \) and \( P_C \) analogously. Show that if
\[
AB' + BC' + CA' = AC' + BA' + CB',
\]
then points \( P, P_A, P_B, \) and \( P_C \) are concyclic.
|
Solution 1 (authorâs).
We will need a couple of lemmas.
**Lemma 3.3.1 (âSparrow lemmaâ)**
Let \( I \) be the incenter of triangle \( ABC \). Point \( U \) lies on ray \( \overrightarrow{AB} \) and point \( V \) lies on ray \( \overrightarrow{CA} \) beyond \( A \) so that \( AU - AV = AB + AC - BC \). Then points \( A, I, U, \) and \( V \) are concyclic.
*Proof.* Let the incircle of triangle \( ABC \) touch sides \( AB \) and \( AC \) at points \( K \) and \( L \), respectively. Since
\[
AK = AL = \frac{1}{2}(AB + AC - BC),
\]
it follows that \( KU = LV \). Hence, right triangles \( IKU \) and \( ILV \) are congruent, and so
\[
\angle AUI = \angle IUK = \angle IVL = \angle AVI.
\]
\(\square\)
**Lemma 3.3.2 (IMO 1979/3)**
Let \( P \) be a common point of the two circles \( \Gamma_1 \) and \( \Gamma_2 \). A variable line \( \ell \) through \( P \) meets \( \Gamma_1 \) and \( \Gamma_2 \) again at points \( T_1 \) and \( T_2 \), respectively. Then, as line \( \ell \) varies, the perpendicular bisectors of all segments \( T_1T_2 \) pass through a constant point.
*Proof.* Let \( D_1 \) and \( D_2 \) be the points diametrically opposite \( P \) in circles \( \Gamma_1 \) and \( \Gamma_2 \), respectively. Then both lines \( D_1T_1 \) and \( D_2T_2 \) are perpendicular to segment \( T_1T_2 \). Thus the perpendicular bisector of segment \( T_1T_2 \) coincides with the mid-line of the strip formed by lines \( D_1T_1 \) and \( D_2T_2 \), and so it always passes through the midpoint of segment \( D_1D_2 \). \(\square\)
We are ready to tackle the problem.
Since
\[
AB' + BC' + CA' = AC' + BA' + CB',
\]
we can find points \( U_A \) and \( V_A \) on ray \( \overrightarrow{PA'} \), \( U_B \) and \( V_B \) on ray \( \overrightarrow{PB'} \), and \( U_C \) and \( V_C \) on ray \( \overrightarrow{PC'} \) simultaneously satisfying all six of the equations
\[
\begin{aligned}
PU_A - PV_B &= PA + PB' - AB', \
PU_A - PV_C &= PA + PC' - AC', \
PU_B - PV_C &= PB + PC' - BC', \
PU_B - PV_A &= PB + PA' - BA', \
PU_C - PV_A &= PC + PA' - CA', \
PU_C - PV_B &= PC + PB' - CB',
\end{aligned}
\]
because any five of the equations implies the sixth one.
Denote by \( I_A, I_B, I_C, J_A, J_B, J_C \) the incenters of \( \triangle BPC', \triangle CPA', \triangle APB', \triangle CPB', \triangle APC', \triangle BPA' \), respectively.
By Lemma 1 applied to triangle \( APB' \), we get that \( P, I_C, U_A, V_A \) are concyclic. Similarly, the other five incenters lie on the analogous circles through \( P \).
Let \( s_A \) be the perpendicular bisector of segment \( I_AJ_A \), let \( t_A \) be the perpendicular bisector of segment \( U_AV_A \), and define lines \( s_B, t_B, s_C, \) and \( t_C \) analogously.
Consider the circumcircles \( (PI_AU_BU_C) \) and \( (PJ_AV_BV_C) \). Applying Lemma 2 to this shows that lines \( s_A, t_B, t_C \) are concurrent at a point \( Q_A \). Analogously, lines \( s_B, t_C, t_A \) meet at some point \( Q_B \) and lines \( s_C, t_A, t_B \) meet at some point \( Q_C \).
Observe that lines \( t_A, t_B, t_C \) are perpendicular to lines \( AA', BB', \) and \( CC' \), respectively, whereas lines \( s_A, s_B, s_C \) are perpendicular to lines \( I_BC I_CB, I_CA I_AC, \) and \( I_AB I_BA \), respectively. Since the latter three lines bisect the pairwise angles between the former three lines, we conclude that lines \( s_A, s_B, s_C \) bisect the interior angles of the triangle formed by lines \( t_A, t_B, \) and \( t_C \).
Therefore, lines \( s_A, s_B, \) and \( s_C \) meet at the incenter \( K \) of triangle \( Q_AQ_BQ_C \). Or, equivalently, all four points \( P, P_A, P_B, \) and \( P_C \) lie on the circle with diameter \( PK \). The solution is complete.
---
Solution 2 (sent by Arjun Gupta).
The proof hinges on the following lemma, which is essentially a restatement of Ptolemyâs theorem.
**Lemma (Trigonometric Ptolemy)**
Let \( \ell_A, \ell_B, \ell_C \) be three lines concurrent at a point \( P \). Let \( \theta_A \) denote value of non-obtuse angle between \( \ell_B \) and \( \ell_C \). Define \( \theta_B, \theta_C \) similarly. Impose a sign convention for lengths on \( \ell_A, \ell_B, \ell_C \) such that among the three rays formed by positive direction through \( P \), no ray lies inside the angle formed by the other two rays. Let \( P_A \) be any point on \( \ell_A \), and define \( \ell_B, \ell_C \) similarly. Then
\[
\sin \theta_A \cdot \overrightarrow{PP_A} + \sin \theta_B \cdot \overrightarrow{PP_B} + \sin \theta_C \cdot \overrightarrow{PP_C} = 0
\]
holds if and only if points \( P, P_A, P_B, P_C \) are concyclic.
*Proof.* Suppose first that \( P, P_A, P_B, P_C \) are concyclic. WLOG points \( P, P_B, P_A, P_C \) lie on a circle in that order. Ptolemyâs Theorem gives
\[
PP_A \cdot PBP_C = PP_B \cdot P_CP_A + PP_C \cdot P_AP_B.
\]
By law of sines,
\[
P_BP_C : P_CP_A : P_AP_B = \sin \theta_A : \sin \theta_B : \sin \theta_C
\]
So we obtain
\[
\sin \theta_A \cdot \overrightarrow{PP_A} = \sin \theta_B \cdot \overrightarrow{PP_B} + \sin \theta_C \cdot \overrightarrow{PP_C}.
\]
Now \( P_A \) lies inside \( \angle P_BP_PC \), so \( \ell_A \) passes through interior of \( \angle P_BP_PC \). This means \( \overrightarrow{PP_B}, \overrightarrow{PP_C} \) have the same sign, WLOG both are non-negative. Now \( \overrightarrow{PP_A} \) must be non-positive, otherwise \( P \) lies inside \( \angle P_AP_BP_C \). It follows
\[
\sin \theta_A (-\overrightarrow{PP_A}) = \sin \theta_B (\overrightarrow{PP_B}) + \sin \theta_C (\overrightarrow{PP_C})
\]
\[\Rightarrow \sin \theta_A \cdot \overrightarrow{PP_A} + \sin \theta_B \cdot \overrightarrow{PP_B} + \sin \theta_C \cdot \overrightarrow{PP_C} = 0.\]
This proves one direction. Now the converse direction just follows from a phantom point argument.
Let \( I_A, J_A \) denote the incenters of triangles \( BPC' \) and \( CPB' \), respectively. Define \( I_B, J_B, I_C, J_C \) similarly. Let \( \ell_A, \ell_B, \ell_C \) denote the lines \( I_AJ_A, I_BJ_B, I_CJ_C \), respectively. Clearly \( P \in \ell_A, \ell_B, \ell_C \).
We will write \( \overrightarrow{AB} \) to denote the directed length of segment \( AB \). Let the lengths on \( \ell_A, \ell_B, \ell_C \) be directed, with lengths \( \overrightarrow{PI_A}, \overrightarrow{PI_B}, \overrightarrow{PI_C} \) being positive.
For a triangle \( PXY \), let \( s(PXY) \) denote the value
\[
\frac{PX + PY - XY}{2}.
\]
Note if \( I \) is incenter of \( \triangle PXY \), then
\[
PI = \frac{s(PXY)}{\cos \frac{\angle XPY}{2}}.
\]
Since
\[
AB' + BC' + CA' = AC' + BA' + CB',
\]
so we obtain
\[
s(PBC') + s(PCA') + s(PAB') = s(PCA') + s(PCB') + s(PAC').
\]
Let \( \theta_A \) be acute angle between lines \( \ell_B \) and \( \ell_C \). Define \( \theta_B, \theta_C \) similarly. Observe that
\[
\theta_A = \frac{\angle BPC}{2} = 90^\circ - \frac{\angle BPC'}{2} = 90^\circ - \frac{\angle CPB'}{2}.
\]
So we obtain
\[
\overrightarrow{PI_A} = \frac{s(PBC')}{\sin \theta_A}, \quad \overrightarrow{PJ_A} = -\frac{s(PCA')}{\sin \theta_A}.
\]
We analogously obtain
\[
\overrightarrow{PP_A} = \frac{\overrightarrow{PI_A} + \overrightarrow{PJ_A}}{2} = \frac{1}{2}\left(\frac{s(PBC') - s(PCA')}{\sin \theta_A}\right),
\]
\[
\overrightarrow{PP_B} = \frac{1}{2}\left(\frac{s(PCA') - s(PCB')}{\sin \theta_B}\right),
\]
\[
\overrightarrow{PP_C} = \frac{1}{2}\left(\frac{s(PAB') - s(PAC')}{\sin \theta_C}\right).
\]
It follows that
\[
\sin \theta_A \cdot \overrightarrow{PP_A} + \sin \theta_B \cdot \overrightarrow{PP_B} + \sin \theta_C \cdot \overrightarrow{PP_C} = 0,
\]
so our earlier Ptolemy-based lemma applies and the problem is solved.
|
As usual, incomplete computational approaches earn partial credits only based on the amount of synthetic progress which is made. No points are awarded for just drawing a diagram or simple observations.
**Solution 1 Rubric**
(i) 2 points are awarded for proving the points \( P, I_C, U_A, V_A \) are concyclic (and the symmetric ones)
(ii) 2 points are awarded for proving \( s_A, t_B, t_C \) meet at \( Q_A \) (and the symmetric ones)
(iii) 3 points for finishing the problem
**Solution 2 Rubric**
(iv) 1 point is awarded for stating the Lemma
(v) 1 point is awarded for proving the Lemma
(vi) 4 points for:
\[
\sin\left(90^\circ + rac{\psi_A}{2}
ight) \cdot \overrightarrow{PP_A} = rac{PB - PC + PC' - PB' - BC' + CB'}{4}
\]
(vii) 1 point is awarded for finishing the problem
**Outside the two solutions**
(viii) If there is any non-trivial linear relation for the tangents from \( P \) to the incircles, 1 point is awarded.
If the student has approaches from both the solutions, they get the maximum of the two possible markings.
|
USEMO-2022-4
|
https://web.evanchen.cc/exams/report-usemo-2022.pdf
|
Let \(ABCD\) be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points \(P, Q, R, S\) lie in the interiors of segments \(AB, BC, CD, DA\), respectively, such that
\[
\angle PDA = \angle PCB, \quad \angle QAB = \angle QDC, \quad \angle RBC = \angle RAD, \quad \text{and} \quad \angle SCD = \angle SBA.
\]
Let \(AQ\) intersect \(BS\) at \(X\), and \(DQ\) intersect \(CS\) at \(Y\). Prove that lines \(PR\) and \(XY\) are either parallel or coincide.
|
Solution 1 (authorâs).
Let \(U = \overline{AD} \cap \overline{BC}\) and \(V = \overline{AB} \cap \overline{CD}\).
**Claim.** We have \(US = UQ\) and \(VP = VR\).
*Proof.* We have
\[
\angle BSA = \angle BAS + \angle SBA = \angle BCD + \angle DCS = \angle BCS
\]
hence
\[
US^2 = UB \cdot UC.
\]
Similarly, \(UQ^2 = UA \cdot UD = UB \cdot UC\). So, \(US = UQ\); similarly \(VP = VR\). â¡
**Claim.** Quadrilateral \(SXQY\) is a kite (with \(SX = SY\) and \(QX = QY\)).
*Proof.* We have
\[
\angle BSQ = \angle USQ - \angle USB = \angle SQU - \angle SCB = \angle QSC
\]
so \(SQ\) bisects \(\angle L BSC\); similarly it bisects \(\angle AQD\). â¡
**Claim.** The internal bisectors of \(\angle U\) and \(\angle V\) are perpendicular.
*Proof.* The angle between these angle bisectors equals
\[
\begin{aligned}
\frac{1}{2}\angle DUC + \angle DAV + \frac{1}{2}\angle BVC &= 90^\circ - \frac{\angle ADC}{2} - \frac{\angle DCB}{2} + \angle BCD + 90^\circ - \frac{\angle ABC}{2} - \frac{\angle DCB}{2} \
&= 90^\circ.
\end{aligned}
\]
As \(\overline{SQ}\) and \(\overline{PR}\) are perpendicular to the internal bisectors of \(\angle U\) and \(\angle V\) by the first claim, so by the third claim \(QS \perp PR\). Meanwhile the second claim gives that \(XY\) is perpendicular to \(\overline{SQ}\), completing the problem. â¡
Solution 2 (due to Nikolai Beluhov).
Let \(E = \overline{AC} \cap \overline{BD}\). Then \(E\) lies on \(\overline{XY}\) by Pappusâs theorem.
**Claim.** Line \(XY\) is the interior bisector of \(\angle AEB\) and \(\angle CED\).
*Proof.* The angle conditions imply that \(X\) and \(Y\) are corresponding points in the two similar triangles \(AEB\) and \(DEC\). Hence, \(\angle AEX = \angle DEY\) and \(\angle BEX = \angle CEY\). Since segments \(EX\) and \(EY\) are collinear, weâre done. â¡
Introduce the points
\[
X' = \overline{BR} \cap \overline{CP}, \quad Y' = \overline{AR} \cap \overline{DP}.
\]
By the same argument as before, line \(\overline{X'EY'}\) is the internal angle bisector of angles \(\angle AED\) and \(\angle BEC\).
**Claim.** Quadrilateral \(PX'RY'\) is a kite (with \(PX' = PY'\) and \(RX' = RY'\)).
*Proof.* Because \(X'\) and \(Y'\) are corresponding points in \(\triangle BEC\) and \(\triangle AED\),
\[
\angle RX'Y' = 180^\circ - \angle BX'E = 180^\circ - \angle AY'E = \angle RY'X',
\]
and so \(RX' = RY'\). Similarly, \(PX' = PY'\). â¡
Thus, \(PR\) is perpendicular to \(\overline{X'EY'}\), hence parallel to the interior bisector of \(\angle LAEB\) and \(\angle LCED\). Together with the first claim, weâre done.
*Remark.* Itâs possible to write up this solution without ever defining \(X'\) and \(Y'\). The idea is to instead prove \(SXQY\) is a kite (which is natural since \(X\) and \(Y\) are already marked) and hence obtain the sentence â\(\overline{SQ}\) is parallel to the internal angle bisector of \(\angle LAED\) and \(\angle LBEC\)â (using the first claim). Then cyclically shift the labels in to get the sentence â\(\overline{PR}\) is parallel to the internal angle bisector of \(\angle LDEC\) and \(\angle LAEB\)â.
|
§3.4b Marking scheme
As usual, incomplete computational approaches earn partial credits only based on the amount of synthetic progress which is made. No points are awarded for just drawing a diagram or simple observations.
There are two major paths a solution can follow:
⢠One which introduces and works with \(\overline{AB} \cap \overline{CD}\) and \(\overline{AD} \cap \overline{BC}\) (the first official solution)
⢠One which works around \(\overline{AC} \cap \overline{BD}\) (the second official solution)
Marks are to be given as the maximum of the scores obtained across either of the approaches.
**Rubric for solution 1**
The following partial items are available and are additive:
⢠1 point for showing that \(US = UQ\) or something similar
⢠3 points for showing that \(SQ\) bisects \(\angle BSC\) and \(\angle AQD\) or any claim analogous to this, and concluding that \(SXQ\) is congruent to \(SYQ\) (or even just directly using this to claim \(XY \perp SQ\)).
⢠2 points for proving that \(PR \perp SQ\)
⢠1 point for completing the solution
For solutions that achieve at most 1 point from the above scheme the following items (additive) are available too:
⢠+1 point for making the correct conjecture that \(SXQY\) is a kite, or equivalently that \(SXQ\) is congruent to \(SYQ\) (with no proof attached)
**Rubric for solution 2**
The following partial items are available and are additive:
⢠0 points for just claiming \(E\) lies on \(XY\) by Pappusâ theorem.
⢠2 points for the claim that \(X\) and \(Y\) are corresponding points in similar triangles \(AEB\) and \(DEC\).
⢠2 points for proving that line \(XY\) bisects \(\angle AEB\) and/or \(\angle CED\).
⢠2 points for showing that \(XY \perp QS\)
⢠1 point for completing the solution
For solutions that achieve at most 2 points from the above scheme the following items (additive) are available too:
⢠+1 point for making the correct conjecture that \(XY \perp SQ\)
⢠+1 point for making the correct conjecture that line \(XY\) bisects \(\angle AEB\) and/or \(\angle CED\)
**Common items for both solutions**
⢠No deduction for configuration issues (such as not using directed angles) or small typos in angle chasing
⢠No deduction for making claims (without written proof) that have reasoning analogous to a claim already proven. (As an example, after showing \(SQ\) bisects \(\angle BSC\) there is no need to prove that \(QS\) bisects \(\angle AQD\)).
⢠Of course, a solution that uses a mixture of both approaches but ends up proving the required condition and is mathematically accurate receives full credit too.
⢠-1 point for skipping the angle chase for showing the angle bisectors of angle \(U\) and angle \(V\) are perpendicular (in case that approach has been taken, and the solution is complete otherwise)
|
USEMO-2022-5
|
https://web.evanchen.cc/exams/report-usemo-2022.pdf
|
Let \( au(n) \) denote the number of positive integer divisors of a positive integer \( n \) (for example, \( au(2022) = 8 \)). Given a polynomial \( P(X) \) with integer coefficients, we define a sequence \( a_1, a_2, \ldots \) of nonnegative integers by setting
\[
a_n = egin{cases}
\gcd(P(n), au(P(n))) & ext{if } P(n) > 0 \
0 & ext{if } P(n) \leq 0
\end{cases}
\]
for each positive integer \( n \). We then say the sequence has \emph{limit infinity} if every integer occurs in this sequence only finitely many times (possibly not at all).
Does there exist a choice of \( P(X) \) for which the sequence \( a_1, a_2, \ldots \) has limit infinity?
|
We claim the answer is no, such \( P \) does not exist.
Clearly we may assume \( P \) is nonconstant with positive leading coefficient. Fix \( P \) and fix constants \( n_0, c > 0 \) such that \( c = P(n_0) > 0 \). We are going to prove that infinitely many terms of the sequence are at most \( c \).
We start with the following lemma.
extbf{Claim} â For each integer \( n \geq 2 \), there exists an integer \( r = r(n) \) such that
- For any prime \( p \) which is at most \( n \), we have \(
u_p(P(r)) =
u_p(c) \).
- We have
\[
c \cdot \prod_{ ext{prime } p \leq n} p \leq r \leq 2c \cdot \prod_{ ext{prime } p \leq n} p.
\]
extit{Proof.} This follows by the Chinese remainder theorem: for each \( p \leq n \) we require \( r \equiv n_0 \pmod{p^{
u_p(c)+1}} \), which guarantees \(
u_p(P(r)) =
u_p(P(n_0)) =
u_p(c) \). Then there exists such an \( r \) modulo \( \prod_{p \leq n} p^{
u_p(c)+1} \) as needed. \( \square \)
Assume for contradiction that all \( a_i \) are eventually larger than \( c \). Take \( n \) large enough that \( n > c \) and \( r = r(n) \) has \( a_r > c \). Then consider the term \( a_r \):
- Using the conditions in the lemma it follows there exists a prime \( p_n > n \) which divides \( a_r = \gcd(P(r), au(P(r))) \) (otherwise \( a_r \), which divides \( P(r) \), is at most \( c \)).
- As \( p_n \) divides \( au(P(r)) \), this forces \( P(r) \) to be divisible by (at least) \( q_n^{p_n - 1} \) for some prime \( q_n \).
- For the small primes \( p \) at most \( n \), we have \(
u_p(P(r)) =
u_p(c) < c < n \leq p_n - 1 \). It follows that \( q_n > n \).
- Ergo,
\[
P(r) \geq q_n^{p_n - 1} > n^n.
\]
In other words, for large enough \( n \), we have the asymptotic estimate
\[
n^n < P(r) = O(1) \cdot r^{\deg P} = O(1) \cdot c^{\deg P} \cdot \prod_{ ext{prime } p \leq n} p^{\deg P} < O(1) \cdot n^{\deg P \cdot \pi(n)}
\]
where \( \pi(n) \) denotes the number of primes less than \( n \). For large enough \( n \) this is impossible since the primes have zero density:
\[
\lim_{n o \infty} rac{\pi(n)}{n} = 0.
\]
extbf{Remark.} For completeness, we outline a short elementary proof that \( \lim_{n o \infty} rac{\pi(n)}{n} = 0 \). For integers \( M > 0 \) define
\[
\delta(M) := \prod_{p \leq M} \left(1 - rac{1}{p}
ight).
\]
Then \( \pi(n) < \delta(M)n + \prod_{p \leq M} p \), so it suffices to check that \( \lim_{M o \infty} \delta(M) = 0 \). But
\[
rac{1}{\delta(M)} = \prod_{p \leq M} \left(1 - rac{1}{p}
ight)^{-1} = \prod_{p \leq M} \left(1 + rac{1}{p} + rac{1}{p^2} + \cdots
ight) \geq 1 + rac{1}{2} + \cdots + rac{1}{M}
\]
which diverges for large \( M \).
|
None of the following items are additive.
(i) 0 points for claiming the answer is no.
(ii) 0 points for solving the problem for linear polynomials.
(iii) 1 point for claiming infinitely many terms of the sequence have value \( c \) for some constant \( c \) in terms of \( P \) whose \(
u_p \)'s are âwell-behavedâ.
(iv) 1 point for creating a useful sequence \( r(n) \) but not finishing.
(v) 2 points for proving that for any constant \( C > 0 \), there exists some prime \( p > C \) and index \( r \), for which \( p \mid a_r \).
(vi) 4 points if, in the previous item, it is also proved that \( r \) is bounded by a reasonable function of \( p \) like \( p^{O(p)} \).
(vii) 6 points for finishing the problem apart from observing primes have zero density.
(viii) No points deducted for stating that \( \pi(n) < Cn \) for big enough \( n \), for any value of \( C \), even without proof.
(ix) No points deducted for small errors caused by \( P(n) \leq 0 \) for finitely many \( n \).
(x) 7 points for a complete solution.
|
USEMO-2022-6
|
https://web.evanchen.cc/exams/report-usemo-2022.pdf
|
Find all positive integers \( k \) for which there exists a \emph{nonlinear} function \( f : \mathbb{Z} o \mathbb{Z} \) which satisfies
\[
f(a) + f(b) + f(c) = rac{f(a-b) + f(b-c) + f(c-a)}{k}
\]
for any integers \( a, b, c \) with \( a + b + c = 0 \).
|
The complete set of solutions is given by
- For \( k = 1 \), \( f(x) \equiv C_1 x + C_2(x mod 2) + C_3 \).
- For \( k = 3 \), \( f(x) \equiv C_1 x + C_2 x^2 \).
- For \( k = 9 \), \( f(x) \equiv C_1 x + C_2 x^4 \).
- For all other \( k \), only \( f(x) \equiv C_1 x \).
Here \( C_1, C_2, C_3 \) are arbitrary integers. We can check they work, so now we just want to show they are the only ones.
We will solve the functional equation for \( f : \mathbb{Z} o \mathbb{C} \), claiming that the above solution set remains the only one. If \( k = 1 \), we can shift by constants to get \( f(0) = 0 \); if \( k
e 1 \) apply \( a = b = c = 0 \) to get \( f(0) = 0 \) anyways. Now note that \( f(x) \equiv x \) is a solution, so we may shift by the identity to assume \( f(-1) = f(1) \).
We will prove in this case, \( f \equiv 0 \) unless \( k = 1, 3, 9 \).
Now plug in \( (a, b, c) = (n+1, -n, -1) \) and \( (a, b, c) = (1, n, -(n+1)) \) gives
\[
f(2n+1) + f(-n+1) + f(-n-2) = k \left(f(n+1) + f(-n) + f(-1)
ight) = k \left(f(-n-1) + f(n) + f(1)
ight).
\]
The last two by induction imply \( f \) is even. Now, by using this and \( (a, b, c) = (n, -n, 0) \) we obtain
\[
f(2n+1) + f(n+2) + f(n-1) = k \left(f(n+1) + f(n) + f(1)
ight) \
f(2n) + 2f(n) = k \left(2f(n) + f(0)
ight) \implies f(2n) = (2k - 2)f(n).
\]
Thus \( f \) is determined recursively by \( f(1) \) (by induction). In particular, if \( f(1) = 0 \) then \( f(n) \equiv 0 \) by induction.
Now, let us assume \( f(1)
e 0 \), and hence by scaling \( f(1) = 1 \). We can then compute:
\[
egin{aligned}
f(1) &= 1 \
f(2) &= 2k - 2 \
f(3) &= k^2 \
f(4) &= 4k^2 - 8k + 4 \
f(5) &= k^3 - 2k^2 + 7k - 5 \
f(6) &= 2k^3 - 2k^2 \
f(7) &= 4k^3 - 6k^2 - 4k + 7 \
f(8) &= 8k^3 - 24k^2 + 24k - 8
\end{aligned}
\]
Plug in \( (a, b, c) = (5, -3, -2) \) and compute \( f(8) + f(1) + f(7) = k(f(5) + f(3) + f(2)) \), which simplifies to give
\[
k^4 - 13k^3 + 39k^2 - 27k = 0 \implies k(k - 1)(k - 3)(k - 9) = 0
\]
so \( k = 1, k = 3, \) or \( k = 9 \). In these cases it is easy to check by induction now that \( f(n) = n mod 2, f(n) = n^2, \) and \( f(n) = n^4 \).
|
In this rubric, a student earns up to 2 points for giving valid constructions, and up to 5 points for proving those solutions are the only ones. These points for the construction are additive with those for the proof.
The construction has three steps:
- Proving \( k = 3 \) works (writing something like: just expand and use \( (a + b + c)^2 = 0 \), counts as a fine proof).
- Proving \( k = 1 \) works (writing something like: check cases according to parity and use \( a + b + c = 0 \), counts as a fine proof).
- Proving \( k = 9 \) works (writing something like: put \( c = -(a + b) \) and just expand the 4th powers, counts as a fine proof).
They are scored as follows:
(i) 1 point for any two of the three constructions
(ii) 2 points for obtaining all three constructions
For the rest of the proof (not additive with each other):
(iii) 0 points for just \( f(0) = 0 \).
(iv) 0 points for relating \( f(x) \) to \( f(-x) \), say by shifting.
(v) 2 points for obtaining a recursion for \( f(0), f(1), \ldots \) in terms of \( k \), such as \( f(2x) = (2k - 2)f(x) \) and \( f(2x + 1) = \ldots \), which in principle allows the computation of any \( f(N) \) for any \( N > 0 \).
In rare cases, itâs possible to award 1 point for more generally showing that if \( f(x) \) has certain âpropertyâ for enough small values of \( |x| \), then \( f(x) \) has that property for every value of \( x \). This is in the same spirit as recursion, but less rigidly defined. The criteria for a âpropertyâ should be consulted with the problem captain, case-by-case.
(vi) 3 points for computing \( f(N) \) in terms of \( k \) for two odd values of \( N \ge 5 \). Itâs okay even if the polynomials is not completely correct (i.e. arithmetic errors).
(vii) 4 points for obtaining a nonzero polynomial in \( k \) which equals 0, say by plugging in \( (a, b, c) = (5, -3, -2) \). Itâs okay even if the polynomials is not completely correct (i.e. arithmetic errors), or the degree of the polynomial is greater than 3.
(viii) 5 points for concluding \( k \in \{1, 3, 9\} \) correctly.
|
USEMO-2023-1
|
https://web.evanchen.cc/exams/report-usemo-2023.pdf
|
A positive integer \( n \) is called \emph{beautiful} if, for every integer \( 4 \leq b \leq 10000 \), the base-\( b \) representation of \( n \) contains the consecutive digits 2, 0, 2, 3 (in this order, from left to right). Determine whether the set of all beautiful integers is finite.
|
We show there are infinitely many beautiful integers. Here are three different approaches.
**Solution 1** (One constructive approach):
We will construct an increasing sequence of positive integers
\[
N_4 < N_5 < N_6 < \cdots
\]
such that for every \( k = 4, 5, \ldots \), the number \( N_k \) contains 2023\(_b\) in every base \( 4 \leq b \leq k \).
This will solve the problem because \( N_{10000}, N_{10001}, \ldots \) will be the requested infinite set.
For the base case, take \( N_4 = 2023_4 \).
For the inductive step, here is one of many valid recipes. We are going to select
\[
N_k = N_{k-1} + c \cdot (k\ell)^e
\]
where the ingredients \( c, \ell, e \) are selected to satisfy:
- \( \ell \) is the product of all primes at most \( k \) which are relatively prime to \( k \) (in particular, \( \gcd(k, \ell) = 1 \));
- \( e \) is large enough that for each \( b = 4, 5, \ldots, k \), the largest power of \( b \) dividing \( (k\ell)^e \) is greater than \( b \cdot N_{k-1} \);
- \( c \) is chosen to satisfy the modular congruence
\[
c \cdot \ell^e \equiv 2k^3 + 0k^2 + 2k + 3 \pmod{k^4}
\]
which is possible since \( \gcd(k^4, \ell^e) = 1 \).
With these ingredients, for all the smaller bases \( 4, 5, \ldots, k-1 \), the ending of \( N_k \) in base-\( b \) is the same as in \( N_{k-1} \) (since \( (k\ell)^e \) is a multiple of a large enough power of \( b \)). On the other hand, weâve embedded 2023\(_k\) into the base-\( k \) representation of \( N_k \), because the coefficients of \( k^{e+3}, k^{e+2}, k^{e+1}, k^e \) in the base-\( k \) representation are exactly 2, 0, 2, 3.
**Solution 2** (An indirect inductive approach):
The goal of this approach is to construct a system of congruences of the form
\[
\begin{aligned}
N &\equiv 2023_4 \cdot 4^{e_4} + t_4 \pmod{4^{e_4+4}} \
N &\equiv 2023_5 \cdot 5^{e_5} + t_5 \pmod{5^{e_5+4}} \
N &\equiv 2023_6 \cdot 6^{e_6} + t_6 \pmod{6^{e_6+4}} \
&\vdots \
N &\equiv 2023_{10000} \cdot 10000^{e_{10000}} + t_{10000} \pmod{10000^{e_{10000}+4}}
\end{aligned}
\]
which has at least one simultaneous solution in \( N \), and where \( 0 \leq t_b < b^{e_b} \). The equation involving \( b^{e_b+4} \) will automatically ensure \( N \) has the desired substring 2023\(_b\), so we concern ourselves only with ensuring the system of equations is consistent.
To do this, we use the following theorem:
**Theorem (Generalized Chinese remainder theorem):**
Fix integers \( a, b, m, n \) with \( m, n > 0 \). The equations
\[
\begin{aligned}
x &\equiv a \pmod{m} \
x &\equiv b \pmod{n}
\end{aligned}
\]
have a simultaneous solution if and only if \( a \equiv b \pmod{\gcd(m,n)} \). Moreover, if there is a solution, that solution is unique modulo \( \operatorname{lcm}(m,n) \).
We proceed by induction, selecting the pairs \( (t_4, e_4), (t_5, e_5), \ldots \) in order. For the base case, it is enough to take \( t_4 = e_4 = 0 \).
Now suppose we have selected pairs up to \( (t_{b-1}, e_{b-1}) \) for some \( b \) and it is time to select \( (t_b, e_b) \). The inductive hypothesis means that the previous equations up to base \( b-1 \) can be collated into a single equivalent equation
\[
N \equiv C \pmod{L} \quad\text{where } L := \operatorname{lcm}(4^{e_4+4}, 5^{e_5+4}, \ldots, (b-1)^{e_{b-1}+4}).
\]
The critical observation is that as long as \( e_b \) is selected large enough so that
\[
b^{e_b} \geq \gcd(L, b^{e_b+4})
\]
then at least one of the choices of \( t_b \) will satisfy
\[
2023_b \cdot b^{e_b} + t_b \equiv C \pmod{\gcd(L, b^{e_b+4})}
\]
which is the compatibility condition needed to ensure
\[
N \equiv 2023_b \cdot b^{e_b} + t_b \pmod{b^{e_b+4}}
\]
can be added to the preceding equations. Since this is obviously possible (by taking \( e_b > \log_b L \)) the induction is complete.
**Solution 3** (Density approach outline):
Itâs enough to prove the following claim:
**Claim â** Fix any integer \( b \geq 4 \). The arithmetic density of nonnegative integers whose base-\( b \) representation does not contain 2023\(_b\) as a contiguous substring is zero.
**Proof.** Think about just the last \( 4n \) digits in base-\( b \), in \( n \) groups of 4. For every complete residue class modulo \( b^{4n} \), the number of base-\( b \) numbers that donât have 2023\(_b\) in their base-\( b \) representation is at most \( (b^4 - 1)^n \).
Consequently, if we have a threshold \( N \), the number of 2023\(_b\)-avoiding numbers in \( \{0, 1, \ldots, N\} \) is bounded above by
\[
(b^4 - 1)^n \cdot \left\lfloor \frac{N}{b^{4n}} \right\rfloor + b^{4n}
\]
and so the density of 2023\(_b\)-avoiding numbers, for large \( N \), is at most
\[
\frac{(b^4 - 1)^n}{b^{4n}} = \left(1 - \frac{1}{b^4}\right)^n.
\]
Since this statement has to hold for any \( n \geq 1 \), the density must be zero.
Thus, it follows that âmostâ positive integers are beautiful! \( \blacksquare \)
|
Marking scheme for inductions and constructions (common approach)
The following remark from the solution packet is key to understanding the rubric:
The essential difficulty in this problem arises from the fact that different bases may share overlapping prime factors, so the Chinese Remainder Theorem does not immediately apply. (Indeed, the solution above does not use the Chinese remainder theorem at all.) To give a concrete example, if \( N \) is an integer whose base-6 representation contains a specific substring, then there are several restrictions on what possible base-9 representations it could have, which are serious enough to limit the control a student has, but not strong enough to actually determine any of the digits. Thatâs why in the inductive proof, there is sort of a âcritical stepâ in which the common prime factors are factored out via \( (k\ell)^e \), after which \( c \) is chosen via modular inverses once the common prime factors have been factored out.
In short, a solution is considered 7â once it can achieve this critical step where, to deal with overlapping prime factors, a large set of common primes are factored out so that a suitable modular inverse can be used in order to complete the proof.
To achieve this benchmark, these steps must be written in sufficient detail to be checked; they should be deduced using explicit equations specifying the necessary thresholds and moduli. For example the official solution reads:
We are going to select \( N_k = N_{k-1} + c \cdot (k\ell)^e \) where the ingredients \( c, \ell, e \) are selected to satisfy:
- \( \ell \) is the product of all primes at most \( k \) which are relatively prime to \( k \) (in particular, \( \gcd(k, \ell) = 1 \));
- \( e \) is large enough that for each \( b = 4, 5, \ldots, k \), the largest power of \( b \) dividing \( (k\ell)^e \) is greater than \( b \cdot N_{k-1} \);
- \( c \) is chosen to satisfy the modular congruence
\[
c \cdot \ell^e \equiv 2k^3 + 0k^2 + 2k + 3 \pmod{k^4}
\]
which is possible since \( \gcd(k^4, \ell^e) = 1 \).
Merely claiming there is \emph{some} congruence of some sort is not sufficient to pass the benchmark.
For solutions that \emph{fail} to achieve this benchmark, the following partial credits are available but \textbf{not additive}:
- 0 points for yes/no answer alone.
- 0 points for just mentioning induction.
- 0 points for base cases like \( b = 4 \).
- 0 points for a solution that only works on coprime moduli. This includes, e.g. showing there is a number which has 2023\(_b\) for every base \( 4 \leq b \leq 2023 \) which is the power of a prime (that is, \( b \in \{4,5,7,8,9,11,13,\ldots,2011\} \)).
- 1 point for showing that the existence of a \emph{single} beautiful number implies the existence of infinitely many, e.g. by adding large powers of 2023!.
- 1 point for a serious induction attempt or construction but which botches the main difficulty mentioned above.
- 2 points for a solution that additionally has the idea in the indirect construction solution of ensuring compatibility by picking a certain integer parameter \( t_i \in \{0,1,\ldots,b^k-1\} \) in a \emph{consecutive range} for a sufficiently large \( k \).
Solutions that pass this benchmark earn:
- 7 points if they are completely correct.
- 6 points for a minor error such as:
- completely omitting the base case of the induction;
- using an integer parameter which is not large enough as stated but could easily be changed to be large enough.
- 5 points for a more serious but non-central flaw in one of the steps of an inductive approach.
The distinction between 5 and 6 will be done more closely by the problem captain.
Marking scheme for arithmetic density approach:
- 0 points for just conjecturing the arithmetic density of beautiful numbers is 1
- 4 points for showing that the number of residue classes modulo \( b^{4n} \) that donât have 2023\(_b\) as a contiguous substring is at most \( (b^4 - 1)^n \).
- 6 points for complete solution modulo minor errors.
- 7 points for complete solution.
|
USEMO-2023-2
|
https://web.evanchen.cc/exams/report-usemo-2023.pdf
|
Each point in the plane is labeled with a real number. Show that there exist two distinct points \( P \) and \( Q \) whose labels differ by less than the distance from \( P \) to \( Q \).
|
Let \( f : \mathbb{R}^2 o \mathbb{R} \) be the labeling, and suppose for contradiction the difference in labels for any points \( P, Q \in \mathbb{R}^2 \) is at least their distance.
egin{quote}
extbf{Claim} â Let \( I \) be a closed interval of length 1. For any \( arepsilon > 0 \), the pre-image
\[
f^{-1}(I) := \{ x \in \mathbb{R}^2 \mid f(x) \in I \}
\]
can be contained inside a set of squares whose total area is at most \( arepsilon \).
\end{quote}
extit{Proof.} Let \( n \ge 1 \). Divide \( I \) into \( n \) closed intervals of length \( frac{1}{n} \). The problem condition implies that the pre-image of each sub-interval is contained inside a square of side length at most \( frac{2}{n} \), and hence with area at most \( \left( frac{2}{n}
ight)^2 = frac{4}{n^2} \). The total area of the squares is thus bounded by \( n \cdot frac{4}{n^2} = frac{4}{n} \). By taking \( n \) large enough that \( arepsilon > 4/n \), weâre done. \( \square \)
Divide the codomain \( \mathbb{R} \) into closed intervals
\[
egin{aligned}
I_1 &= [0,1] \
I_2 &= [-1,0] \
I_3 &= [1,2] \
I_4 &= [-2,-1] \
I_5 &= [2,3] \
I_6 &= [-3,-2] \
& dots
\end{aligned}
\]
By the claim, the pre-image \( f^{-1}(I_k) \) could be contained inside squares whose total area is at most, say, \( 10^{-k} \). So the entire pre-image \( f^{-1}(\mathbb{R}) \) could be contained inside squares whose total area is at most \( \sum_{k \ge 1} 10^{-k} = frac{1}{9} \), which is finite. But this is absurd, since \( f^{-1}(\mathbb{R}) = \mathbb{R}^2 \).
extbf{Remark} (Generalization of the problem). One can imagine the same problem with the target condition modified to
\[
|f(P) - f(Q)| < |PQ|^c
\]
for a general \( c > 0 \); the present problem is \( c = 1 \).
For \( 0 < c < 2 \), the above proof works equally well. For \( c = 2 \), we provide a construction where the statement is no longer true.
It suffices to find a function \( f : \mathbb{C} o \mathbb{R} \) such that
\[
|f(z_2) - f(z_1)| > |z_2 - z_1|^2
\]
for all complex numbers \( z_1 \) and \( z_2 \).
Let \( g(z) : [0,1] + [0,1]i o [0,1] \) be the inverse Hilbert curve. The preimage of any interval \( \left[ frac{n}{2^{2k}}, frac{n+1}{2^{2k}}
ight] \) is a square of side length \( frac{1}{2^k} \) that is adjacent to the preimage of \( \left[ frac{n+1}{2^{2k}}, frac{n+2}{2^{2k}}
ight] \). This means the preimage of any length \( \ell \) interval is contained in a width \( 4\sqrt{\ell} \) square. This means
\[
8\sqrt{|g(z_2) - g(z_1)|} > |z_2 - z_1|,
\]
implying that some sufficiently large multiple of \( g(z) \), say \( h(z) \), satisfies the desired inequality over its domain.
To extend the domain of this solution to all complex numbers, partition the complex plane into countably many unit squares, copy \( h(z) \) onto each unit square, and space the images of each unit square sufficiently far apart on the real number line.
|
For all solutions, the following are not awarded marks:
⢠Proving the statement for all continuous labellings.
⢠Assuming there exists a labelling \( f : \mathbb{R}^2 o \mathbb{R} \) such that \( |f(x) - f(y)| \ge |x - y| \) for the sake of contradiction.
⢠Proving \( f \) is injective.
For solutions similar to the official solution, the following items are available but not additive:
⢠1 point for examining the pre-image of an interval.
⢠1 point for examining a 2-dimensional set (as opposed to a lattice or a countable number of lines) and mentioning area.
⢠2 points for proving \( f^{-1}([a,b]) \) is contained in a set of area \( C(b-a)^2 \) for some constant \( C \). For reference, this set will (usually) be either a square or a disk.
⢠5 points for proving the above quadratic bound extbf{and} introducing a harmonic covering of \( \mathbb{R} \).
⢠5 points for proving \( f^{-1}([a,b]) \) is contained in a set of area \( arepsilon \) for any \( arepsilon > 0 \).
⢠7 points for a complete solution.
Solutions using the advanced theory of Lipschitz functions are scored as follows, with all marks being additive:
⢠+1 point for showing \( f^{-1} \) is 1-Lipschitz extbf{and} surjective on its domain \( S \).
⢠+2 points for a suitable extension of \( f^{-1} \) to \( g : \mathbb{R}^2 o \mathbb{R}^2 \). Graders should look out for implicit assumptions that \( S \) is composed of a finite number of intervals/dense or not dense/Borel/Lebesgue-measurable/etc.
⢠+1 point for covering the preimage of \( f^{-1}(\mathbb{R}) \) with a countable number of discs with finite area, awarded extbf{only if} the contestant hits the above 2 points.
For all solutions which are incomplete with errors, the following deductions apply and are all additive:
⢠-1 point for only partitioning \( \mathbb{R}^+ \) into intervals.
⢠-1 point for mentions of the âareaâ of \( f^{-1}([a,b]) \) (instead of that of a superset).
|
USEMO-2023-3
|
https://web.evanchen.cc/exams/report-usemo-2023.pdf
|
Canmoo is trying to do constructions, but doesnât have a ruler or compass. Instead, Canmoo has a device that, given four distinct points \( A, B, C, P \) in the plane, will mark the isogonal conjugate of \( P \) with respect to triangle \( ABC \), if it exists. Show that if two points are marked on the plane, then Canmoo can construct their midpoint using this device, a pencil for marking additional points, and no other tools.
(Recall that the \emph{isogonal conjugate} of \( P \) with respect to triangle \( ABC \) is the point \( Q \) such that lines \( AP \) and \( AQ \) are reflections around the bisector of \( ngle BAC \), lines \( BP \) and \( BQ \) are reflections around the bisector of \( ngle CBA \), lines \( CP \) and \( CQ \) are reflections around the bisector of \( ngle ACB \). Additional points marked by the pencil can be assumed to be in general position, meaning they donât lie on any line through two existing points or any circle through three existing points.)
|
We assume Canmoo can mark points in arbitrarily general position.
We first prove two claims showing that reflection around a point is possible. We will only use the second claim in what proceeds (so with the second claim proven, we can forget about the first one.)
extbf{Claim â} Given any three points \( X, B, C \), Canmoo can construct the reflection of \( X \) over \( BC \).
extit{Proof.} Pick another point \( P \), and let \( Y \) be the isogonal conjugate of \( X \) in \( riangle PBC \), and \( Z \) the isogonal conjugate of \( X \) in \( riangle YBC \). Then
\[ ngle CBX = ngle YBP = ngle ZBC, \quad ext{and} \quad ngle BCX = ngle YCP = ngle ZCB. \]
Thus, \( Z \) is the reflection of \( X \) over \( BC \). \(\square\)
extbf{Claim â} Given any two points \( A \) and \( B \), Canmoo can reflect \( A \) over \( B \).
extit{Proof.} Pick another point \( C \). Let \( D \) and \( F \) denote the reflections of \( A \) and \( C \) over \( BC \) and \( AB \). Then reflect \( D \) over \( CF \) to \( X \), and \( B \) over \( DX \) to \( Y \), so that \( BY \perp AB \). Finally, the reflection of \( A \) over \( BY \) is the reflection of \( A \) over \( B \). \(\square\)
extbf{Claim â} Given any three points \( A, B, C \), Canmoo can construct the point \( P \) such that \( AP \parallel BC \) and \( ngle APC = 90^\circ \).
extit{Proof.} WLOG assume \( ngle B
e 90^\circ \). (If not, replace \( B \) with its reflection over \( C \).) Reflect \( A \) over \( B \) and \( C \) to get \( B' \) and \( C' \), respectively. Reflect \( B' \) over \( C \) to get \( S \), so \( AB'C'S \) is a parallelogram which is not a rectangle. Since \( AP \parallel B'C' \), the isogonal of \( AP \) with respect to \( ngle B'AC' \) is the tangent to the circumcenter of \( riangle ABC \) at \( A \); and hence the isogonal conjugate of \( S \) is \( T := AA \cap C'C' \), the intersection of the tangents at \( A \) and \( C' \), as shown below.
Now take the isogonal conjugate of \( T \) with respect to \( riangle ABC \). It still lies on line \( AS \), but because \( ngle TCA = 90^\circ \) we have \( ngle PCB = 90^\circ \) too, hence \( ngle APC = 90^\circ \). So \( P \) is exactly the point desired. \(\square\)
extbf{Claim â} Given any three points \( A, B, C \), Canmoo can construct the foot from \( A \) to \( BC \).
extit{Proof.} Construct the point \( P \) in the third claim, and WLOG assume none of the angles in the problem are \( 90^\circ \) (otherwise, reflect \( B \) or \( C \) over each other), so \( P \) is distinct from \( A \). Then apply the construction again to \( BAP \) to get the desired foot. \(\square\)
We are ready to tackle the main construction. Let \( B \) and \( C \) be the initial two given points, and pick a third point \( A \) and assume \( ngle A
e 90^\circ \). As in the proof of the third claim, reflect \( A \) over \( B, C \) to get \( B' \) and \( C' \), and reflect \( B' \) over \( C \) to \( S \).
Then let \( D \) be the reflection of \( S \) across \( C' \), hence \( AB'DC' \) is a parallelogram. The isogonal conjugate of \( D \) with respect to \( riangle AB'C' \) is therefore the intersection \( X \) of the tangents to \( (AB'C') \) at \( B' \) and \( C' \). Then taking the foot from \( X \) to \( B'C' \) gives the foot from \( M \) to \( B'C' \).
In particular, \( ABMC \) is a parallelogram, and so again we may take the isogonal conjugate of \( M \) with respect to \( riangle ABC \) to obtain the point \( Y = BB \cap CC \) which is the intersection of the tangents to \( B \) and \( C \) at \( (ABC) \). Finally, taking the foot from \( Y \) to \( BC \) gives the desired midpoint.
|
For the most part, solutions could be read on a case-by-case basis, because not many solutions have nontrivial progress. We standardize the following extbf{non-additive} benchmarks:
- extbf{1 point} for showing \( X \) can be reflected over \( YZ \).
- extbf{0 points} for showing \( X \) can be reflected over a point \( Y \).
- extbf{3 points} for being able to obtain the feet of the altitudes in a triangle
- extbf{3 points} for being able to obtain the circumcenters of triangles.
|
USEMO-2023-4
|
https://web.evanchen.cc/exams/report-usemo-2023.pdf
|
Let \( ABC \) be an acute triangle with orthocenter \( H \). Points \( A_1, B_1, C_1 \) are chosen in the interiors of sides \( BC, CA, AB \), respectively, such that \( riangle A_1B_1C_1 \) has orthocenter \( H \). Define \( A_2 = \overrightarrow{AH} \cap B_1C_1 \), \( B_2 = \overrightarrow{BH} \cap C_1A_1 \), and \( C_2 = \overrightarrow{CH} \cap A_1B_1 \). Prove that triangle \( A_2B_2C_2 \) has orthocenter \( H \).
|
Solution 1: Power of a point solution, by Nikolai Beluhov.
In this solution, all lengths are signed. Let \( riangle DEF \) be the orthic triangle of \( riangle ABC \), and \( riangle D_1E_1F_1 \) be the orthic triangle of \( riangle A_1B_1C_1 \). We define two common quantities, through power of a point:
\[ k := HA \cdot HD = HB \cdot HE = HC \cdot HF. \]
\[ k_1 := HA_1 \cdot HD_1 = HB_1 \cdot HE_1 = HC_1 \cdot HF_1. \]
Because quadrilateral \( A_2D_1DA_1 \) is concyclic (with circumdiameter \( \overline{A_1A_2} \)), by power of a point, we get
\[ HA_2 \cdot HD = HD_1 \cdot HA_1 = k_1 \Rightarrow HA_2 = rac{k_1}{HD} = rac{k_1}{k} \cdot HA. \]
Since \( k_1/k \) is fixed, a symmetric argument now gives
\[ rac{HA_2}{HA} = rac{HB_2}{HB} = rac{HC_2}{HC} = rac{k_1}{k}. \]
Therefore, \( H \) is the center of a homothety mapping \( riangle A_2B_2C_2 \) to \( riangle ABC \). In particular, it is also the orthocenter of \( riangle A_2B_2C_2 \).
Solution 2: Authorâs ratio-based solution.
We are going to prove:
**Claim** â We have \( B_2C_2 \parallel BC \).
Refer to the diagram. Note that
\[
rac{C_1A_2}{A_2B_1} = rac{[AC_1H]}{[AB_1H]} = rac{AC_1 \cdot d(H, AB)}{AB_1 \cdot d(H, AC)} = rac{ rac{AC_1}{HC}}{ rac{AB_1}{HB}} = rac{HB}{HC} \cdot rac{\sin ngle AB_1C_1}{\sin ngle AC_1B_1} = rac{HB}{HC} \cdot rac{\sin ngle CB_1A_1}{\sin ngle C_1HA_1} = rac{[HB A_1]}{[HC A_1]} = rac{BA_1}{A_1C}.
\]
Similarly, \( rac{A_1B_2}{B_2C_1} = rac{CB_1}{B_1A} \) and \( rac{B_1C_2}{C_2A_1} = rac{AC_1}{C_1B} \). Hence,
\[
[BB_2C] = [BC_1C] \cdot rac{B_2A_1}{C_1A_1} = [BAC] \cdot rac{B_2A_1}{C_1A_1} \cdot rac{C_1B}{AB} = [ABC] \cdot rac{B_1C}{AC} \cdot rac{C_1B}{AB}.
\]
Similarly, \( [BC_2C] \) also equals this quantity, so \( B_2C_2 \parallel BC \) and \( A_2H \perp B_2C_2 \).
Repeating this we see that \( H \) is the orthocenter of \( riangle A_2B_2C_2 \), as wanted.
Solution 3: Alternative finish to first solution by inversion.
Define \( D, D_1, E, E_1, F \) and \( F_1 \) as in the first solution. As before, the quadrilateral \( A_2D_1DA_1 \) is concyclic (with circumdiameter \( A_1A_2 \)).
Now, rather than using power of a point, consider the negative inversion with center \( H \) and radius \( \sqrt{HA_1 \cdot HD} \). Since \( A_2D_1DA_1 \) is cyclic and similarly, this inversion maps \( E \) to \( B_2 \) and \( F \) to \( C_2 \). Hence the circle \( (EHF) \) maps to the line \( B_2C_2 \).
But the circle \( (EHF) \) is symmetric with respect to line \( DH \) (the center of the circle is the midpoint of \( \overline{AH} \)), so the circle \( (EHF) \) must map to a line perpendicular to \( DH \). It follows that \( B_2C_2 \perp DH \), or that \( A_2H \perp B_2C_2 \), as needed.
Solution 4: Alternative finish to first solution using Reimâs theorem.
From the cyclic quadrilaterals \( A_2D_1DA_1 \) and similar in the first solution we also have
\[ HC_2 \cdot HF = HC_1 \cdot HF_1 = HB_1 \cdot HE_1 = HB_2 \cdot HE. \]
From this it follows that \( B_2C_2EF \) is cyclic.
We also know \( BCEF \) is cyclic and hence by Reimâs Theorem we get that \( B_2C_2 \parallel BC \), which implies the result.
|
General rules:
- As usual, incomplete computational approaches earn partial credits only based on the amount of synthetic progress which is made.
- No points are awarded for just drawing a diagram or simple observations.
- No points are deducted for configuration issues (such as not using directed angles) and minor typos.
- If the student has approaches from more than one of the solutions, they get the maximum of all possible markings.
Items worth zero points:
0 points for stating the problem is equivalent to prove that \( AB \parallel A_2B_2 \), or that \( ABC \) and \( A_2B_2C_2 \) are homothetic with center \( H \).
**Power of a point solution, by Nikolai Beluhov**
The following partial items are available and are **additive**:
(a) +1 point: Proving that \( A_1, D, A_2 \) and \( D_1 \) lie on a circle.
(b) +1 point: Showing it is sufficient to prove \( rac{HA_2}{HA} \) is constant (or \( rac{HA_2}{HA} = rac{HB_2}{HB} = rac{HC_2}{HC} \), or anything else identical).
**Authorâs ratio-based solution**
The following partial items are available and are **additive**:
(a) +2 points: Proving that \( rac{C_1A_2}{A_2B_1} = rac{BA_1}{A_1C} \).
(b) +1 point: Showing it is sufficient to prove \( [BB_2C] = [BC_2B] \) (or that the distances from \( B_2 \) and \( C_2 \) to \( BC \) are equal).
**Inversion (third solution)**
The following partial items are available and are **additive**, though note that altogether they form a complete solution:
(a) +1 point: Proving that \( A_1, D, A_2 \) and \( D_1 \) lie on a circle.
(b) +1 point: Stating that the line \( DH \) and the circle \( (EHF) \) are orthogonal (or that the center of \( (EHF) \) lies on \( DH \)).
(c) +1 point: Reducing the problem (e.g. by inversion) to showing that the line \( DH \) and the circle \( (EHF) \) are orthogonal (or that the center of \( (EHF) \) lies on \( DH \)).
**Angle chase with power of a point (fourth solution)**
The following partial items are available, but only (a) and (b) are **additive**. Note that altogether (b) and (c) form a complete solution.
(a) 1 point: Proving that \( A_1, D, A_2 \) and \( D_1 \) lie on a circle.
(b) 1 point: Showing it is sufficient to prove that \( B_2, C_2, E \) and \( F \) lie on a circle.
(c) 2 points: Proving that \( B_2, C_2, E \) and \( F \) lie on a circle.
Any complete solution is worth **7 points**.
|
USEMO-2023-5
|
https://web.evanchen.cc/exams/report-usemo-2023.pdf
|
Let \( n \geq 2 \) be an integer. A cube of size \( n \times n \times n \) is dissected into \( n^3 \) unit cubes. A nonzero real number is written at the center of each unit cube so that the sum of the \( n^2 \) numbers in each slab of size \( 1 \times n \times n \), \( n \times 1 \times n \), or \( n \times n \times 1 \) equals zero. (There are a total of \( 3n \) such slabs, forming three groups of \( n \) slabs each such that slabs in the same group are parallel and slabs in different groups are perpendicular.)
Could it happen that some plane in three-dimensional space separates the positive and the negative written numbers? (The plane should not pass through any of the numbers.)
|
We show this can never happen.
Suppose, for the sake of contradiction, that such a plane \( lpha \) did exist. Let \( Oxyz \) be a Cartesian coordinate system whose origin \( O \) lies in \( lpha \) and whose axes are parallel to the edges of our cube. Let the equation of \( lpha \) in this coordinate system be \( ax + by + cz = 0 \). Without loss of generality, all positive written numbers lie in the half-space \( ax + by + cz > 0 \) relative to \( lpha \) and all negative written numbers lie in the half-space \( ax + by + cz < 0 \) relative to \( lpha \).
For all \( i \in \{1, 2, \dots, n^3\} \), let \( r_i \) be the number written at point \( (x_i, y_i, z_i) \). Then for all \( i \) we have that \( (ax_i + by_i + cz_i)r_i > 0 \).
Therefore,
\[
egin{aligned}
0 &= a \cdot 0 + b \cdot 0 + c \cdot 0 \
&= \sum_x ax \cdot \left( \sum_{i : x_i = x} r_i
ight) + \sum_y by \cdot \left( \sum_{i : y_i = y} r_i
ight) + \sum_z cz \cdot \left( \sum_{i : z_i = z} r_i
ight) \
&= \sum_i ax_i r_i + \sum_i by_i r_i + \sum_i cz_i r_i \
&= \sum_i (ax_i + by_i + cz_i) r_i \
&> 0.
\end{aligned}
\]
We have arrived at a contradiction. The solution is complete.
**Remark.** The so-called Farkas lemma guarantees that if there wasnât a valid labeling, then we can combine the given inequalities to obtain a contradiction; therefore there is a sense in which an inequality-based solution like the one above is a _priori_ promised to exist if the answer is no.
**Remark (Neil Kolekar).** More generally, if \( P, Q, R \) are nonconstant polynomials with real coefficients, then from \( 0 = \sum_{i : x_i = x} r_i \) being true for each \( x \) (and similarly), we should have
\[
0 = \sum_x P(x) \cdot \left( \sum_{i : x_i = x} r_i
ight) + \sum_y Q(y) \cdot \left( \sum_{i : y_i = y} r_i
ight) + \sum_z R(z) \cdot \left( \sum_{i : z_i = z} r_i
ight).
\]
This means the same proof would work in general when the hyperplanes \( ax + by + cz = 0 \) are replaced by more general surfaces of the form
\[
P(x) + Q(y) + R(z) = 0.
\]
|
In what follows, we let \( f(x, y, z) \) be the number written at \( (x, y, z) \) with \( 1 \leq x, y, z \leq n \). Also let the required plane be \( ax + by + cz - d = 0 \).
**Partial items for 0+ solutions**
The following partial items apply for incomplete solutions and are additive.
- **0 points** for claiming the answer
- **0 points** for making the false claim that a plane cannot pass through all slabs
- **0 points** for solving the problem for \( n = 2 \) or \( n = 3 \)
- **+2 points** for considering an expression of the form
\[ \sum x f(x, y, z) \]
and showing that it is equal to zero.
- **+2 points** for considering the product \( (ax + by + cz - d)f(x, y, z) \) or something similar
**Complete solutions**
In case of a complete solution:
- **No deduction** for not explicitly considering at most one of the following cases:
1. \( f(x, y, z) \) and \( ax + by + cz - d \) have the _same_ sign
2. \( f(x, y, z) \) and \( ax + by + cz - d \) have _opposite_ sign
|
USEMO-2023-6
|
https://web.evanchen.cc/exams/report-usemo-2023.pdf
|
Let \( n \geq 2 \) be a fixed integer.
(a) Determine the largest positive integer \( m \) (in terms of \( n \)) such that there exist complex numbers \( r_1, \ldots, r_n \), not all zero, for which
\[
\prod_{k=1}^n (r_k + 1) = \prod_{k=1}^n (r_k^2 + 1) = \cdots = \prod_{k=1}^n (r_k^m + 1) = 1.
\]
(b) For this value of \( m \), find all possible values of
\[
\prod_{k=1}^n (r_k^{m+1} + 1).
\]
|
For part (a) the answer is \( m = 2^n - 2 \); for part (b) the answer is \( 2^n \).
**Construction for (a).** For \( m = 2^n - 2 \), fix \( \omega := \exp\left(\frac{2\pi i}{2^n - 1}\right) \) and set
\[
r_j = \omega^{2^j}, \quad j = 1, 2, \ldots, n, \quad m = 2^n - 2.
\]
We can expand to see that the
\[
\prod_{k=1}^n (r_k^j + 1) = \sum_{j=0}^{2^n - 1} \omega^j = 1, \quad j = 1, 2, \ldots, 2^n - 2.
\]
**Bound for (a).** It remains to show when \( m = 2^n - 1 \) we must have \( r_1 = \cdots = r_n = 0 \). For each nonempty subset \( S \) of \( \{1, \ldots, n\} \), define
\[
\Pi_S := \prod_{k \in S} r_k.
\]
Then the problem condition, when expanded, states that
\[
0 = -1 + \prod_{k=1}^n (r_k^j + 1) = \sum_{\varnothing \neq S \subseteq \{1, \ldots, n\}} \Pi_S^j, \quad j = 1, 2, \ldots, 2^n - 1.
\]
But view the \( \Pi_S \) as \( 2^n - 1 \) variables. Then the first \( 2^n - 1 \) power sums of the \( \Pi_S \) all vanish, and hence (say, by Newton sums) it follows that every \( \Pi_S \) must be zero. In particular, all the variables are zero as well.
**Proof for (b).** Fix \( m = 2^n - 2 \) and
\[
\omega := \exp\left(\frac{2\pi i}{2^n - 1}\right).
\]
We are going to prove that:
**Claim â** Every \( r_i \) is a power of \( \omega \).
**Proof.** If we apply Newton sums as we did before, with the identity
\[
0 = -1 + \prod_{k=1}^n (r_k^j + 1) = \sum_{\varnothing \neq S \subseteq \{1, \ldots, n\}} \Pi_S^j, \quad j = 1, 2, \ldots, 2^n - 1,
\]
then we get that all the elementary symmetric polynomials in \( \Pi_S \) vanish, except for the last one. In other words, the polynomial identity
\[
X^{2^n - 1} - c = \prod_{\varnothing \neq S \subseteq \{1, \ldots, n\}} (X - \Pi_S)
\]
should hold for some \( c \). We saw already (in (a)) that if \( c = 0 \) then \( r_i = 0 \) for all \( i \), so assume \( c \neq 0 \), and that \( r_i \neq 0 \) for all \( i \).
Let \( \lambda \) be any complex \( (2^n - 1) \)th root of \( c \). Then the factorization of the left-hand polynomial over \( \mathbb{C} \) is given exactly by
\[
X^{2^n - 1} - c = \prod_{j=0}^{2^n - 2} (X - \lambda \omega^j).
\]
Hence, we have the equality of unordered sets
\[
\left\{ \Pi_S \mid \varnothing \neq S \subseteq \{1, \ldots, n\} \right\} = \left\{ \lambda \omega^j \mid 0 \leq j \leq 2^n - 2 \right\}.
\]
In particular,
\[
r_1 = \frac{r_1 r_2}{r_2} = \frac{\Pi_{\{1,2\}}}{\Pi_{\{2\}}} = \frac{\lambda \omega^{n_{12}}}{\lambda \omega^{n_2}} = \omega^{n_{12} - n_2}
\]
for some integers \( n_{12} \) and \( n_2 \) (whose values are unimportant); so \( r_1 \) is a \( (2^n - 1) \)th root of unity, and similarly so is every \( r_i \).
Hence
\[
\prod_{i=1}^n (r_i^{m} + 1) = \prod_{i=1}^n (r_i^{2^n - 1} + 1) = \prod_{i=1}^n (1 + 1) = \prod_{i=1}^n 2 = 2^n
\]
is the only possible value of the product requested in (b).
**Remark.** An interesting problem is to characterize all \( (r_1, \ldots, r_n) \). The author has not solved that yet.
|
Part (a) will be graded out of 5 points and part (b) will be graded out of 2 points. The scores of the two parts will be added for the final score out of 7.
**Scoring for (a)**
For incomplete solutions, the following items are available but **not additive**:
- 1 point for the correct answer of \( m = 2^n - 2 \) in part (a)
- 0 points for expanding the product and rewriting it as a sum.
- 2 points for expanding the product and rewriting it as a sum, using Newton sums or Vandermonde Determinant to argue that the answer is at most \( 2^n - 2 \).
- 2 points for a correct construction and answer.
- 5 points for completely solving part (a).
For complete solutions, the following deductions apply, and are **additive**:
- -1 point for wrong answer.
- -1 point for another mistake.
**Scoring for (b)**
- 1 point for correct answer
- 2 points for correct solution.
|
USEMO-2024-1
|
https://web.evanchen.cc/exams/report-usemo-2024.pdf
|
There are 1001 stacks of coins \( S_1, S_2, \ldots, S_{1001} \). Initially, stack \( S_k \) has \( k \) coins for each \( k = 1, 2, \ldots, 1001 \). In an operation, one selects an ordered pair \( (i, j) \) of indices \( i \) and \( j \) satisfying \( 1 \le i < j \le 1001 \) subject to two conditions:
- The stacks \( S_i \) and \( S_j \) must each have at least one coin.
- The ordered pair \( (i, j) \) must not have been selected in any previous operation.
Then, if \( S_i \) and \( S_j \) have \( a \) coins and \( b \) coins, respectively, one removes \( \gcd(a, b) \) coins from each stack.
What is the maximum number of times this operation could be performed?
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The answer is \( 500 \cdot 501 = 250500 \). Our solution is split into two parts.
### Construction
Firstly, we will give a valid construction. We start by performing operations \( (1001, 1000), (1001, 999), \ldots, (1001, 1) \), in order. By induction, at each step \( (1001, j) \), \( S_{1001} \) will have \( j + 1 \) coins and thus, since \( \gcd(j + 1, j) = 1 \), one coin will be removed from each stack. At the end of this process, 1000 operations will have been performed. Then stack \( S_{1001} \) will have one coin; we discard it. The remaining (nonempty) stacks will have \( 1, 2, \ldots, 999 \) coins, and no operation will have been performed between any of them. Thus we can repeat this process, performing operations with the 999-coin stack and the rest of the stacks in descending order.
Repeating this process until all the stacks have been discarded, we perform
\[
1000 + 998 + \cdots + 2 = 500 \cdot 501
\]
operations, as desired.
### Proof of bound
To prove this is the maximum number of operations we can perform, we bound the total number of operations. The stacks \( S_1, \ldots, S_{500} \) can only participate in at most
\[
1 + \cdots + 500 = rac{500 \cdot 501}{2}
\]
operations (since each operation removes at least one coin from them). The remaining 501 stacks can only perform
\[
inom{501}{2} = rac{500 \cdot 501}{2}
\]
operations between themselves, since each pair can only perform the operation once. Thus, in total, we can perform at most \( 500 \cdot 501 \) operations.
|
The solution is split into two parts: the lower bound (construction), worth **4 points**, and the upper bound, worth **3 points**. These parts are completely additive.
In general, minor errors will be worth a deduction, but please message the channel when you find any that are not included in the rubric so that we can add them to the rubric for a deduction. Errors purely in arithmetic (even in the final answer), including incorrect summation of an arithmetic series, will **not** merit any deduction.
### Lower Bound
The following items are available (here \( n = 1001 \)):
- **1 point** for any construction that achieves \( \Omega(n^2) \) moves.
- **2 points** for any construction that achieves \( n^2/4 - O(n) \) moves.
- **4 points** for a correct lower bound construction (achieving \( 500 \cdot 501 = (n^2 - 1)/4 \) moves).
Up to 1 point may be deducted for constructions that are correct as stated but for which insufficient justification is provided that the construction works. (For example, in the case of the construction of the official solution, no justification would be required, since the fact that it works is obvious enough not to require justification.)
### Upper Bound
The following items are available:
- **1 point** for any correct upper bound that is **strictly less than** \( \left\lfloor rac{1001 \cdot 1002}{4}
ight
floor = 250750 \).
- **3 points** for a correct upper bound (of \( 500 \cdot 501 = (n^2 - 1)/4 \)).
|
USEMO-2024-2
|
https://web.evanchen.cc/exams/report-usemo-2024.pdf
|
Let \( k \) be a fixed positive integer. For each integer \( 1 \le i \le 4 \), let \( x_i \) and \( y_i \) be positive integers such that their least common multiple is \( k \). Suppose that the four points \( (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) \) are the vertices of a non-degenerate rectangle in the Cartesian plane. Prove that \( x_1x_2x_3x_4 \) is a perfect square.
|
It suffices to prove that \( v_p(x_1x_2x_3x_4) \) is even for each prime \( p \mid k \). Since the four points form a rectangle, we have
\[
egin{aligned}
x_1 + x_3 &= x_2 + x_4 ag{3.1} \
y_1 + y_3 &= y_2 + y_4 ag{3.2} \
(x_1 - x_3)^2 + (y_1 - y_3)^2 &= (x_2 - x_4)^2 + (y_2 - y_4)^2 ag{3.3} \
x_2x_4 - x_1x_3 &= y_1y_3 - y_2y_4 ag{3.4}
\end{aligned}
\]
Let \( v_p(k) = m \). For each \( 1 \le i \le 4 \), we have
\[
\max\left(v_p(x_i), v_p(y_i)
ight) = m.
\]
We split into cases.
- **Case 1.** If \( v_p(x_i) = m \) for three \( i \) then \( v_p(x_i) = m \) for the fourth by (3.1), so \( 2 \mid v_p(x_1x_2x_3x_4) = 4m \).
- **Case 2.** If \( v_p(y_i) = m \) for three \( i \) then \( v_p(y_i) = m \) for the fourth by (3.2). By (3.4) we have \( p^{2m} \mid x_2x_4 - x_1x_3 \). We now use the fact that \( v_p(x \pm y) = \min(v_p(x), v_p(y)) \) whenever \( v_p(x)
e v_p(y) \). Using the contrapositive, \( v_p(x_1x_3), v_p(x_2x_4) \le 2m \) implies \( v_p(x_1x_3) = v_p(x_2x_4) \) and hence \( 2 \mid v_p(x_1x_2x_3x_4) \).
- **Case 3.** Otherwise, \( v_p(x_i) = m \) for exactly two \( i \). If these \( i \) are consecutive (cyclically), for example \( i = 1, 2 \) without loss of generality, then from (3.1) we have \( p^m \mid x_3 - x_4 \). Since \( v_p(x_3), v_p(x_4) \le m \), we have \( v_p(x_3) = v_p(x_4) \) and hence \( 2 \mid v_p(x_1x_2x_3x_4) \). If these \( i \) are not consecutive, for example \( i = 1, 3 \) without loss of generality, then from (3.1) we have \( p^m \mid x_2 + x_4 \), and we can finish using the same argument as in the consecutive case.
**Remark.** There are rectangles which satisfy the hypothesis, for instance \( (a,b), (a,ab), (ab,ab), (ab,b) \) where \( \gcd(a,b) = 1 \).
|
Recall the four equations
\[
egin{aligned}
x_1 + x_3 &= x_2 + x_4 \
y_1 + y_3 &= y_2 + y_4 \
(x_1 - x_3)^2 + (y_1 - y_3)^2 &= (x_2 - x_4)^2 + (y_2 - y_4)^2 \
x_2x_4 - x_1x_3 &= y_1y_3 - y_2y_4
\end{aligned}
\]
which were numbered (3.1), (3.2), (3.3), and (3.4) in the provided solution.
The following are awarded marks, with items being *non-additive*:
- **2 points** for writing down (3.1), (3.2), and (3.4). The condition \( (y_2 - y_1)(y_4 - y_1) = -(x_2 - x_1)(x_4 - x_1) \) or similar in lieu of (3.4) is accepted but (3.3) or similar is not.
- **2 points** for solving the problem in at least one of Case 2 and Case 3, irrespective of whether the conditions (3.1), (3.2), and (3.4) are written down.
- **5 points** for both of the above items.
- **7 points** for a complete solution.
|
USEMO-2024-3
|
https://web.evanchen.cc/exams/report-usemo-2024.pdf
|
Let ABC be an acute triangle with incenter I. Two distinct points P and Q are chosen on the circumcircle of ABC such that
\[ngle API = ngle AQI = 45^\circ.\]
Lines PQ and BC meet at S. Let H denote the foot of the altitude from A to BC. Prove that \(ngle AHI = ngle ISH.\)
|
We give three solutions.
### Solution 1 (via Tebault circles from the author)
Construct the Tebault circles \(\omega_1\) and \(\omega_2\) which are tangent to (ABC), side BC, and cevian AH.
The key claim is that P and Q coincide with the tangency points of the circles we just drew:
> **Claim** â Points P and Q are the tangency points of \(\omega_i\) and (ABC).
*Proof.* Let \(\omega_1\) touch BC at T and (ABC) at \(P'\). We will show that \(ngle AP'I = 45^\circ\).
Let W be the arc midpoint of BC not containing A and let \(L = AW \cap BC\). It is well known that W, T, P are collinear and (by âshooting lemmaâ)
\[WT \cdot WP = WL \cdot WA = WI^2 = WB^2.\]
Hence we get
\[ngle ITH = ngle ITW - ngle LTW = -ngle P'IW - ngle LTW.\]
From \(AP'TL\) concyclic we also have
\[-ngle P'IW - ngle LTW = -ngle P'IW - ngle IAP' = ngle AP'I.\]
It is left to remember that \(ngle ITH = \pm 45^\circ\) because IT must be parallel to an angle bisector of \(ngle AHB\) by the properties of Tebault circles.
Similarly, \(ngle AQ'I = 45^\circ\). So P and Q coincide with \(P'\) and \(Q'\) in some order. \(\square\)
> **Claim** â Point S is the center of the positive homothety which maps \(\omega_1\) to \(\omega_2\).
*Proof.* This follows by Monge theorem on \(\omega_1, \omega_2\) and (ABC). \(\square\)
Let D be the tangency point of incircle \(\omega\) of triangle ABC with BC with antipode \(D'\) and let \(\ell\) be another tangent form S to \(\omega_1, \omega_2\) (which is a tangent to \(\omega\) as well because of the Tebault circles properties). Let \(K = \ell \cap AH\).
> **Claim** â SHIK is cyclic.
*Proof.* Because of the Tebault circles property, the intersection of cevian and the second tangent lies on the tangent to \(\omega\) at \(D'\). In our case, it follows \(KD'\) is tangent to \(\omega\) and thus parallel to the line BC. As a consequence, SI is parallel to another angle bisector of \(ngle SKD'\). Hence, \(ngle KIS = 90^\circ = ngle KHS\) as desired. \(\square\)
From the third lemma we may conclude that
\[ngle KHI = ngle KSI = ngle ISH,\]
as desired.
> **Remark.** For basic properties about Tebault circles contains the relevant facts (regrettably, it is in Russian only): https://geometry.ru/articles/protasovtebo.pdf.
### Solution 2 (by reviewers)
Let \(A^*\) be the antipodal point of A on (ABC).
> **Lemma 3.3.1** \(ngle AHI = ngle A^*IA.\)
*Proof.* Let \(I_A\) be the excenter opposite A. Since \(ABH \sim AA^*C\) and \(ACH \sim AA^*B\), we get that \(AA^* \cdot AH = AB \cdot AC\). Since \(ABI \sim AI_AC\) and \(ACI \sim AI_AB\), similarly \(AI \cdot AI_A = AB \cdot AC\). But also \(ngle A^*AI = ngle HAI\), and we conclude that \(AIH \sim AA^*I_A\). So \(ngle AHI = ngle AI_AI^*\) Let \(M_A\) be the midpoint of \(II_A\). Then as \(M_A\) is on (ABC), we know that \(AM_A \perp MA^*\), and so \(A^*\) is on the perpendicular bisector of \(II_A\). Thus \(ngle AI_AA^* = ngle A^*IA\), as desired.
Note that \(AIH \sim AA^*I_A\) can be seen simply by taking the \(\sqrt{AB \cdot AC}\)-inversion at A as well. \(\square\)
From now on, we will be proving \(ngle A^*IA = ngle (SI, BC).\)
Let K, L be the second intersections of PI, QI and (ABC), respectively. Then the angle condition is equivalent to that KL is the perpendicular bisector of \(AA^*\).
> **Lemma 3.3.2** SI passes through the circumcenter \(O'\) of triangle \(AIA^*\).
*Proof.* Let U and V be the midpoints of the arcs AB and AC of (ABC). Then as U, V are both on the perpendicular bisector of AI, we know that \(O' = UV \cap KL\). It thus suffices to show that \(UV, KL, SI\) are concurrent.
Let IS intersect (ABC) at X, Y. Then
\[ (S, I; X, Y) \overset{ngle}{=} (B, U; X, Y) \overset{ngle}{=} (I, UV \cap SI; X, Y) \]
and
\[ (S, I; X, Y) \overset{ngle}{=} (Q, K; X, Y) \overset{ngle}{=} (I, KL \cap SI; X, Y). \]
Thus \(UV \cap SI = KL \cap SI\), as desired. \(\square\)
The rest is a straightforward angle chase. We know that
\[ ngle (SI, BC) = ngle SO'K + ngle (KL, BC) = ngle IA^*A + ngle A^*AI = ngle A^*IA, \]
as desired.
### Solution 3 (by Hans Yu)
Let \(I_A\) be the A-excenter. Now let X be the second intersection of (PIQ) and (BIC). Since S is the radical center of the circles (PIQ), (BIC) and (ABC), we see that S, X, I are collinear.
> **Claim 3.3.3** â It suffices to show that \(I_AHX\) are collinear.
*Proof of the claim.* Suppose that \(I_AHX\) are collinear. It is well-known that BC bisects \(I_AHI\) (say, by harmonicity of \((I, I_A; A, BC \cap AI)\)). Therefore \(ngle IHA = ngle AHX = 90^\circ - ngle (XI, AH) = 90^\circ - ngle (SI, AH) = ngle HSI\), as desired. Here we used that \(ngle HXI = ngle I_AXI = 90^\circ\). \(\square\)
Now let AH intersect (ABC) again at D.
> **Claim 3.3.4** â It suffices to show that \(AI_AXD\) are concyclic.
*Proof of the claim.* Suppose that \(AI_AXD\) are concyclic, then H = BC \cap AD is the radical center of (ABCD), (AIXD) and (BICI_A). Hence H is also on \(I_AX\), and we are done by Claim 3.3.4. \(\square\)
PI, QI intersect the circumcircle of ABC again at \(P', Q'\), and let \(A^*\) be the antipodal point of A on the circumcircle of ABC. Then since \(ngle APP' = ngle API = 45^\circ\) and so is \(ngle AQQ'\), we see that \(AP'A^*Q'\) is a square.
Let \(M_A\) be the midpoint of \(II_A\). Let O be the circumcenter of ABC. Then we can see that \(M_A\) is the midpoint of arc \(A^*D\) as well: to see this, note that \(ngle DAA^* = ngle OMA = ngle M_AAO = ngle M_AA^*\).
Let \(O'\) be the circumcenter of \(AIA^*\), and let DI intersect (ABC) again at \(D'\).
> **Claim 3.3.5** â It suffices to show that \(M_AD' \perp O'D'\).
*Proof of the claim.* Suppose that \(M_AD' \perp O'D'\). Let N be the midpoint of AI. Then \(M_AND'O'\) are concyclic.
Consider the inversion at I sending A to \(M_A\). This inversion fixes the circumcircle of ABC. It sends \(D'\) to D and N to \(I_A\) as \(IAI = 2MAI\). Now to see where X is sent to, note that (PQI) is sent to the line \(P'Q'\), which is the perpendicular bisector of \(AA^*\). Moreover, (BIAC) is sent to the line through N perpendicular to AI, which is just the perpendicular bisector of AI. Thus X is sent to the circumcenter of \(AIA^*\), which is \(O'\). As a consequence, \(O'\) is sent to X, and so the circle \(MAND'O'\) is sent to the circle \(AIADX\), and we are done by Claim 2. \(\square\)
To finish off, we will show that \(M_AD' \perp O'D'\). Let Z be on ID' such that \(A^*Z \parallel D'M_A\). Then \(ngle IZA^* = ngle ID'M_A = ngle DD'M_A = ngle M_AD'A^* = ngle IAA^*\), showing that Z is on the circumcircle of \(AIA^*\). As a consequence, \(O'\) is on the perpendicular bisector of \(A^*Z\). However, since \(D'ZA^* = ngle DZA^* = ngle DZM_A = ngle M_AD'A^* = ngle ZA^*D'\), we have that \(D'\) is on the perpendicular bisector of \(A^*Z\) as well. This shows that \(O'D' \perp A^*Z \parallel M_AD'\), as desired.
|
For all solutions, the following are not awarded marks:
- Rephrasing the angle condition in terms of PI \cap (ABC) and QI \cap (ABC).
- Swapping \(ngle AHI\) with some other angles, even if they are used in the official solutions.
For solutions not using Tebault circles, the following items are not additive:
- **2 points** Showing that SI passes through the circumcenter of \(AIA^*\). Alternatively, show that SI, the perpendicular bisector of AI and the perpendicular bisector of AA^* are concurrent.
Note: Points are still awarded if SI is replaced by some other two points that clearly lie on S, I, the perpendicular bisector of AI is replaced by the line connecting two points that are clearly on the perpendicular bisector, or the perpendicular bisector of AA^* is similarly replaced.
- **2 points** Show that if SI passes through the circumcenter of \(AIA^*\), then the statement holds true.
- **7 points** Complete solution.
For solutions using Tebault circles, the following items are additive:
- **+1 point** Show that P, Q are tangency points of the Tebault circles to the circumcircle.
- **+1 point** Show that S is the center of homothety of the two Tebault circles.
- **+3 points** Construct K and show that SHIK are concyclic.
- **+2 points** Finishing the solution.
|
USEMO-2024-4
|
https://web.evanchen.cc/exams/report-usemo-2024.pdf
|
Find all sequences \( a_1, a_2, \ldots \) of nonnegative integers such that for all positive integers \( n \), the polynomial
\[
1 + x^{a_1} + x^{a_2} + \cdots + x^{a_n}
\]
has at least one integer root. (Here \( x^0 = 1. \))
|
The only answer is \( a_1 = 1 \) and \( a_2 = a_3 = \cdots = 0 \).
Itâs clear that this works because for each \( n \), the requested integer root is \( x = -n \). We now prove this is the only solution.
In general, let
\[ F_n(x) := 1 + x^{a_1} + \cdots + x^{a_n}. \]
> **Claim** â Let \( p \) be any prime. Then
\[ F_{p-1}(-(p-1)) = 0. \]
*Proof.* Let \( -r \) be the integer root of \( F_{p-1} \), for \( r > 0 \). From
\[ F_{p-1}(1) = p ext{ and } F_{p-1}(-r) = 0 \implies 1 + r \mid p \]
we conclude that \( r = p - 1 \) (since \( p \) is prime), as needed. \( \square \)
We continue to focus on \( F_{p-1}(-(p - 1)) = 0 \) for any prime \( p \), that is,
\[
1 + \sum_{i=1}^{p-1} (-1)^{a_i} (p - 1)^{a_i} = 0.
\]
The idea is that the big terms are *way* too big. Indeed, set \( M := \max(a_1, \ldots, a_{p-1}) \) and assume that \( M \) occurs for \( k \ge 1 \) indices among \( \{a_1, \ldots, a_{p-1}\} \). Hence in the displayed sum, there are \( k \) terms equal to \( (-1)^M (p - 1)^M \). Hence
\[
k \cdot (p - 1)^M = \left| 1 + \sum_{i : a_i < M} (-1)^{a_i} (p - 1)^{a_i}
ight| \le 1 + (p - 1 - k) \cdot (p - 1)^{M-1}
\]
which gives
\[
1 \ge (p - 1)^{M - 1} [k \cdot (p - 1) - (p - 1 - k)] = (p - 1)^{M - 1} [(k - 1)(p - 1) + k].
\]
This could only happen if \( k = 1 \) and \( M = 1 \). In other words, for any prime \( p \), the terms \( (a_1, \ldots, a_{p-1}) \) consist of a single 1 and all other 0âs.
In particular, for \( p = 2 \) we have \( a_1 = 1 \). Hence \( a_2 = a_3 = \cdots = 0 \) as desired.
|
For all solutions, the following are not awarded marks:
- Getting the correct answer with no explanation.
- Showing that all roots are negative.
For correct solutions:
- **7 points** for a complete solution that shows \( a_1 = 1 \) and \( a_{i > 1} = 0 \) is the only possible solution.
Solutions that are not complete will get the *maximum* points any item for partial credit may award them, and these points are **not additive**:
- **+2 points** for noting that if \( n \) is one less than a prime, \( 1 - p \) must be the integer root.
- **+4 points** for using Extremal Principle and taking the largest value of \( a_i \) into account.
- **+4 points** for showing that \( P_{p-1}(0) = r Q_{p-1}(0) \) so \( p - 1 \mid P_{p-1}(0) \).
- **+5 points** for setting the absolute value equation up in either of the two previous cases.
- **+5 points** for showing that \( \sum_{i=0}^{p-1} x^{a_i} = x + (p - 1) \).
For all solutions which are incomplete with errors, the following deductions apply and are all additive. An incomplete solution can only get a deduction if it applies for the *complete* portion:
- **-1 point** for not checking that \( a_1 = 1 \) and \( a_{i > 1} = 0 \) is indeed valid.
- **-1 point** for not indicating in any way that we check all primes \( p \).
|
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