# USA IMO TST 2016 Solutions <br> United States of America - IMO Team Selection Tests <br> Evan Chen《陳誼廷》 <br> 60 ${ }^{\text {th }}$ IMO 2016 Hong Kong 

## Contents

0 Problems ..... 2
1 Solutions to Day 1 ..... 3
1.1 USA TST 2016/1, proposed by Maria Monks ..... 3
1.2 USA TST 2016/2, proposed by Evan Chen ..... 4
1.3 USA TST 2016/3, proposed by Mark Sellke ..... 6
2 Solutions to Day 2 ..... 8
2.1 USA TST 2016/4, proposed by Iurie Boreico ..... 8
2.2 USA TST 2016/5, proposed by Zilin Jiang ..... 9
2.3 USA TST 2016/6, proposed by Ivan Borsenco ..... 10

## §0 Problems

1. Let $S=\{1, \ldots, n\}$. Given a bijection $f: S \rightarrow S$ an orbit of $f$ is a set of the form $\{x, f(x), f(f(x)), \ldots\}$ for some $x \in S$. We denote by $c(f)$ the number of distinct orbits of $f$. For example, if $n=3$ and $f(1)=2, f(2)=1, f(3)=3$, the two orbits are $\{1,2\}$ and $\{3\}$, hence $c(f)=2$.
Given $k$ bijections $f_{1}, \ldots, f_{k}$ from $S$ to itself, prove that

$$
c\left(f_{1}\right)+\cdots+c\left(f_{k}\right) \leq n(k-1)+c(f)
$$

where $f: S \rightarrow S$ is the composed function $f_{1} \circ \cdots \circ f_{k}$.
2. Let $A B C$ be a scalene triangle with circumcircle $\Omega$, and suppose the incircle of $A B C$ touches $B C$ at $D$. The angle bisector of $\angle A$ meets $B C$ and $\Omega$ at $K$ and $M$. The circumcircle of $\triangle D K M$ intersects the $A$-excircle at $S_{1}, S_{2}$, and $\Omega$ at $T \neq M$. Prove that line $A T$ passes through either $S_{1}$ or $S_{2}$.
3. Let $p$ be a prime number. Let $\mathbb{F}_{p}$ denote the integers modulo $p$, and let $\mathbb{F}_{p}[x]$ be the set of polynomials with coefficients in $\mathbb{F}_{p}$. Define $\Psi: \mathbb{F}_{p}[x] \rightarrow \mathbb{F}_{p}[x]$ by

$$
\Psi\left(\sum_{i=0}^{n} a_{i} x^{i}\right)=\sum_{i=0}^{n} a_{i} x^{x^{i}} .
$$

Prove that for nonzero polynomials $F, G \in \mathbb{F}_{p}[x]$,

$$
\Psi(\operatorname{gcd}(F, G))=\operatorname{gcd}(\Psi(F), \Psi(G))
$$

4. Let $\sqrt{3}=1 . b_{1} b_{2} b_{3} \cdots(2)$ be the binary representation of $\sqrt{3}$. Prove that for any positive integer $n$, at least one of the digits $b_{n}, b_{n+1}, \ldots, b_{2 n}$ equals 1 .
5. Let $n \geq 4$ be an integer. Find all functions $W:\{1, \ldots, n\}^{2} \rightarrow \mathbb{R}$ such that for every partition $[n]=A \cup B \cup C$ into disjoint sets,

$$
\sum_{a \in A} \sum_{b \in B} \sum_{c \in C} W(a, b) W(b, c)=|A||B||C| .
$$

6. Let $A B C$ be an acute scalene triangle and let $P$ be a point in its interior. Let $A_{1}$, $B_{1}, C_{1}$ be projections of $P$ onto triangle sides $B C, C A, A B$, respectively. Find the locus of points $P$ such that $A A_{1}, B B_{1}, C C_{1}$ are concurrent and $\angle P A B+\angle P B C+$ $\angle P C A=90^{\circ}$.

## §1 Solutions to Day 1

## §1.1 USA TST 2016/1, proposed by Maria Monks

Available online at https://aops.com/community/p5679356.

## Problem statement

Let $S=\{1, \ldots, n\}$. Given a bijection $f: S \rightarrow S$ an orbit of $f$ is a set of the form $\{x, f(x), f(f(x)), \ldots\}$ for some $x \in S$. We denote by $c(f)$ the number of distinct orbits of $f$. For example, if $n=3$ and $f(1)=2, f(2)=1, f(3)=3$, the two orbits are $\{1,2\}$ and $\{3\}$, hence $c(f)=2$.

Given $k$ bijections $f_{1}, \ldots, f_{k}$ from $S$ to itself, prove that

$$
c\left(f_{1}\right)+\cdots+c\left(f_{k}\right) \leq n(k-1)+c(f)
$$

where $f: S \rightarrow S$ is the composed function $f_{1} \circ \cdots \circ f_{k}$.

Most motivated solution is to consider $n-c(f)$ and show this is the transposition distance. Dumb graph theory works as well.

## §1.2 USA TST 2016/2, proposed by Evan Chen

Available online at https://aops.com/community/p5679361.

## Problem statement

Let $A B C$ be a scalene triangle with circumcircle $\Omega$, and suppose the incircle of $A B C$ touches $B C$ at $D$. The angle bisector of $\angle A$ meets $B C$ and $\Omega$ at $K$ and $M$. The circumcircle of $\triangle D K M$ intersects the $A$-excircle at $S_{1}, S_{2}$, and $\Omega$ at $T \neq M$. Prove that line $A T$ passes through either $S_{1}$ or $S_{2}$.

We present an angle-chasing solution, and then a more advanced alternative finish.
【 First solution (angle chasing). Assume for simplicity $A B<A C$. Let $E$ be the contact point of the $A$-excircle on $B C$; also let ray $T D$ meet $\Omega$ again at $L$. From the fact that $\angle M T L=\angle M T D=180^{\circ}-\angle M K D$, we can deduce that $\angle M T L=\angle A C M$, meaning that $L$ is the reflection of $A$ across the perpendicular bisector $\ell$ of $B C$. If we reflect $T, D$, $L$ over $\ell$, we deduce $A, E$ and the reflection of $T$ across $\ell$ are collinear, which implies that $\angle B A T=\angle C A E$.

Now, consider the reflection point $E$ across line $A I$, say $S$. Since ray $A I$ passes through the $A$-excenter, $S$ lies on the $A$-excircle. Since $\angle B A T=\angle C A E, S$ also lies on ray $A T$. But the circumcircles of triangles $D K M$ and $K M E$ are congruent (from $D M=E M$ ), so $S$ lies on the circumcircle of $\triangle D K M$ too. Hence $S$ is the desired intersection point.
![](https://cdn.mathpix.com/cropped/2024_11_19_2f35297903aafb6226b5g-04.jpg?height=1212&width=783&top_left_y=1390&top_left_x=642)

Second solution (advanced). It's known that $T$ is the touch-point of the $A$-mixtilinear incircle. Let $E$ be contact point of $A$-excircle on $B C$. Now the circumcircles of $\triangle D K M$ and $\triangle K M E$ are congruent, since $D M=M E$ and the angles at $K$ are supplementary. Let $S$ be the reflection of $E$ across line $K M$, which by the above the above comment lies on the circumcircle of $\triangle D K M$. Since $K M$ passes through the $A$-excenter, $S$ also lies on the $A$-excircle. But $S$ also lies on line $A T$, since lines $A T$ and $A E$ are isogonal (the mixtilinear cevian is isogonal to the Nagel line). Thus $S$ is the desired intersection point.

【 Authorship comments. This problem comes from an observation of mine: let $A B C$ be a triangle, let the $\angle A$ bisector meet $\overline{B C}$ and $(A B C)$ at $E$ and $M$. Let $W$ be the tangency point of the $A$-mixtilinear excircle with the circumcircle of $A B C$. Then $A$ Nagel line passed through a common intersection of the circumcircle of $\triangle M E W$ and the $A$-mixtilinear incircle.

This problem is the inverted version of this observation.

## §1.3 USA TST 2016/3, proposed by Mark Sellke

Available online at https://aops.com/community/p5679392.

## Problem statement

Let $p$ be a prime number. Let $\mathbb{F}_{p}$ denote the integers modulo $p$, and let $\mathbb{F}_{p}[x]$ be the set of polynomials with coefficients in $\mathbb{F}_{p}$. Define $\Psi: \mathbb{F}_{p}[x] \rightarrow \mathbb{F}_{p}[x]$ by

$$
\Psi\left(\sum_{i=0}^{n} a_{i} x^{i}\right)=\sum_{i=0}^{n} a_{i} x^{p^{i}} .
$$

Prove that for nonzero polynomials $F, G \in \mathbb{F}_{p}[x]$,

$$
\Psi(\operatorname{gcd}(F, G))=\operatorname{gcd}(\Psi(F), \Psi(G))
$$

Observe that $\Psi$ is also a linear map of $\mathbb{F}_{p}$ vector spaces, and that $\Psi(x P)=\Psi(P)^{p}$ for any $P \in \mathbb{F}_{p}[x]$. (In particular, $\Psi(1)=x$, not 1 , take caution!)

『 First solution (Ankan Bhattacharya). We start with:
Claim - If $P \mid Q$ then $\Psi(P) \mid \Psi(Q)$.
Proof. Set $Q=P R$, where $R=\sum_{i=0}^{k} r_{i} x^{i}$. Then

$$
\Psi(Q)=\Psi\left(P \sum_{i=0}^{k} r_{i} x^{i}\right)=\sum_{i=0}^{k} \Psi\left(P \cdot r_{i} x^{i}\right)=\sum_{i=0}^{k} r_{i} \Psi(P)^{p^{i}}
$$

which is divisible by $\Psi(P)$.
This already implies

$$
\Psi(\operatorname{gcd}(F, G)) \mid \operatorname{gcd}(\Psi(F), \Psi(G))
$$

For the converse, by Bezout there exists $A, B \in \mathbb{F}_{p}[x]$ such that $A F+B G=\operatorname{gcd}(F, G)$, so taking $\Psi$ of both sides gives

$$
\Psi(A F)+\Psi(B G)=\Psi(\operatorname{gcd}(F, G))
$$

The left-hand side is divisible by $\operatorname{gcd}(\Psi(F), \Psi(G))$ since the first term is divisible by $\Psi(F)$ and the second term is divisible by $\Psi(G)$. So $\operatorname{gcd}(\Psi(F), \Psi(G)) \mid \Psi(\operatorname{gcd}(F, G))$ and noting both sides are monic we are done.

【 Second solution. Here is an alternative (longer but more conceptual) way to finish without Bezout lemma. Let $\beth \subseteq \mathbb{F}_{p}[x]$ denote the set of polynomials in the image of $\Psi$, thus $\Psi: \mathbb{F}_{p}[x] \rightarrow \beth$ is a bijection on the level of sets.

Claim - If $A, B \in \beth$ then $\operatorname{gcd}(A, B) \in \beth$.

Proof. It suffices to show that if $A$ and $B$ are monic, and $\operatorname{deg} A>\operatorname{deg} B$, then the remainder when $A$ is divided by $B$ is in $\beth$. Suppose $\operatorname{deg} A=p^{k}$ and $B=x^{p^{k-1}}-$ $c_{2} x^{p^{k-2}}-\cdots-c_{k}$. Then

$$
\begin{aligned}
x^{p^{k}} & \equiv\left(c_{2} x^{p^{k-2}}+c_{3} x^{p^{k-3}}+\cdots+c_{k}\right)^{p} \quad(\bmod B) \\
& \equiv c_{2} x^{p^{k-1}}+c_{3} x^{p^{k-2}} \cdots+c_{k} \quad(\bmod B)
\end{aligned}
$$

since exponentiation by $p$ commutes with addition in $\mathbb{F}_{p}$. This is enough to imply the conclusion. The proof if $\operatorname{deg} B$ is smaller less than $p^{k-1}$ is similar.

Thus, if we view $\mathbb{F}_{p}[x]$ and $\beth$ as partially ordered sets under polynomial division, then gcd is the "greatest lower bound" or "meet" in both partially ordered sets. We will now prove that $\Psi$ is an isomorphism of the posets. We have already seen that $P|Q \Longrightarrow \Psi(P)| \Psi(Q)$ from the first solution. For the converse:

Claim - If $\Psi(P) \mid \Psi(Q)$ then $P \mid Q$.

Proof. Suppose $\Psi(P) \mid \Psi(Q)$, but $Q=P A+B$ where $\operatorname{deg} B<\operatorname{deg} P$. Thus $\Psi(P) \mid$ $\Psi(P A)+\Psi(B)$, hence $\Psi(P) \mid \Psi(B)$, but $\operatorname{deg} \Psi(P)>\operatorname{deg} \Psi(B)$ hence $\Psi(B)=0 \Longrightarrow$ $B=0$ 。

This completes the proof.
Remark. In fact $\Psi: \mathbb{F}_{p}[x] \rightarrow \beth$ is a ring isomorphism if we equip $\beth$ with function composition as the ring multiplication. Indeed in the proof of the first claim (that $P|Q \Longrightarrow \Psi(P)|$ $\Psi(Q)$ ) we saw that

$$
\Psi(R P)=\sum_{i=0}^{k} r_{i} \Psi(P)^{p^{i}}=\Psi(R) \circ \Psi(P)
$$

## §2 Solutions to Day 2

## §2.1 USA TST 2016/4, proposed by lurie Boreico

Available online at https://aops.com/community/p6368201.

## Problem statement

Let $\sqrt{3}=1 . b_{1} b_{2} b_{3} \cdots(2)$ be the binary representation of $\sqrt{3}$. Prove that for any positive integer $n$, at least one of the digits $b_{n}, b_{n+1}, \ldots, b_{2 n}$ equals 1 .

Assume the contrary, so that for some integer $k$ we have

$$
k<2^{n-1} \sqrt{3}<k+\frac{1}{2^{n+1}} .
$$

Squaring gives

$$
\begin{aligned}
k^{2}<3 \cdot 2^{2 n-2} & <k^{2}+\frac{k}{2^{n}}+\frac{1}{2^{2 n+2}} \\
& \leq k^{2}+\frac{2^{n-1} \sqrt{3}}{2^{n}}+\frac{1}{2^{2 n+2}} \\
& =k^{2}+\frac{\sqrt{3}}{2}+\frac{1}{2^{2 n+2}} \\
& \leq k^{2}+\frac{\sqrt{3}}{2}+\frac{1}{16} \\
& <k^{2}+1
\end{aligned}
$$

and this is a contradiction.

## §2.2 USA TST 2016/5, proposed by Zilin Jiang

Available online at https://aops.com/community/p6368185.

## Problem statement

Let $n \geq 4$ be an integer. Find all functions $W:\{1, \ldots, n\}^{2} \rightarrow \mathbb{R}$ such that for every partition $[n]=A \cup B \cup C$ into disjoint sets,

$$
\sum_{a \in A} \sum_{b \in B} \sum_{c \in C} W(a, b) W(b, c)=|A||B \| C| .
$$

Of course, $W(k, k)$ is arbitrary for $k \in[n]$. We claim that $W(a, b)= \pm 1$ for any $a \neq b$, with the sign fixed. (These evidently work.)

First, let $X_{a b c}=W(a, b) W(b, c)$ for all distinct $a, b, c$, so the given condition is

$$
\sum_{a, b, c \in A \times B \times C} X_{a b c}=|A||B||C| .
$$

Consider the given equation with the particular choices

- $A=\{1\}, B=\{2\}, C=\{3,4, \ldots, n\}$.
- $A=\{1\}, B=\{3\}, C=\{2,4, \ldots, n\}$.
- $A=\{1\}, B=\{2,3\}, C=\{4, \ldots, n\}$.

This gives

$$
\begin{aligned}
X_{123}+X_{124}+\cdots+X_{12 n} & =n-2 \\
X_{132}+X_{134}+\cdots+X_{13 n} & =n-2 \\
\left(X_{124}+\cdots+X_{12 n}\right)+\left(X_{134}+\cdots+X_{13 n}\right) & =2(n-3) .
\end{aligned}
$$

Adding the first two and subtracting the last one gives $X_{123}+X_{132}=2$. Similarly, $X_{123}+X_{321}=2$, and in this way we have $X_{321}=X_{132}$. Thus $W(3,2) W(2,1)=$ $W(1,3) W(3,2)$, and since $W(3,2) \neq 0$ (clearly) we get $W(2,1)=W(3,2)$.

Analogously, for any distinct $a, b, c$ we have $W(a, b)=W(b, c)$. For $n \geq 4$ this is enough to imply $W(a, b)= \pm 1$ for $a \neq b$ where the choice of sign is the same for all $a$ and $b$.

Remark. Surprisingly, the $n=3$ case has "extra" solutions for $W(1,2)=W(2,3)=$ $W(3,1)= \pm 1, W(2,1)=W(3,2)=W(1,3)=\mp 1$.

Remark (Intuition). It should still be possible to solve the problem with $X_{a b c}$ in place of $W(a, b) W(b, c)$, because we have about far more equations than variables $X_{a, b, c}$ so linear algebra assures us we almost certainly have a unique solution.

## §2.3 USA TST 2016/6, proposed by Ivan Borsenco

Available online at https://aops.com/community/p6368189.

## Problem statement

Let $A B C$ be an acute scalene triangle and let $P$ be a point in its interior. Let $A_{1}, B_{1}$, $C_{1}$ be projections of $P$ onto triangle sides $B C, C A, A B$, respectively. Find the locus of points $P$ such that $A A_{1}, B B_{1}, C C_{1}$ are concurrent and $\angle P A B+\angle P B C+\angle P C A=$ $90^{\circ}$.

In complex numbers with $A B C$ the unit circle, it is equivalent to solving the following two cubic equations in $p$ and $q=\bar{p}$ :

$$
\begin{aligned}
(p-a)(p-b)(p-c) & =(a b c)^{2}(q-1 / a)(q-1 / b)(q-1 / c) \\
0 & =\prod_{\text {cyc }}(p+c-b-b c q)+\prod_{\text {сус }}(p+b-c-b c q)
\end{aligned}
$$

Viewing this as two cubic curves in $(p, q) \in \mathbb{C}^{2}$, by Bézout's Theorem it follows there are at most nine solutions (unless both curves are not irreducible, but it's easy to check the first one cannot be factored). Moreover it is easy to name nine solutions (for $A B C$ scalene): the three vertices, the three excenters, and $I, O, H$. Hence the answer is just those three triangle centers $I, O$ and $H$.

Remark. On the other hand it is not easy to solve the cubics by hand; I tried for an hour without success. So I think this solution is only feasible with knowledge of algebraic geometry.

Remark. These two cubics have names:

- The locus of $\angle P A B+\angle P B C+\angle P C A=90^{\circ}$ is the McCay cubic, which is the locus of points $P$ for which $P, P^{*}, O$ are collinear.
- The locus of the pedal condition is the Darboux cubic, which is the locus of points $P$ for which $P, P^{*}, L$ are collinear, $L$ denoting the de Longchamps point.
Assuming $P \neq P^{*}$, this implies $P$ and $P^{*}$ both lie on the Euler line of $\triangle A B C$, which is possible only if $P=O$ or $P=H$.

Some other synthetic solutions are posted at https://aops.com/community/c6h1243902p6368189.