USAMO/md/en-USAMO-1996-notes.md

USAMO 1996 Solution Notes

Evan Chen《陳誼廷》

15 April 2024

This is a compilation of solutions for the 1996 USAMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.

These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let $\mathbb{R}$ denote the set of real numbers" are typically omitted entirely.

Corrections and comments are welcome!

Contents

0 Problems ..... 2 1 Solutions to Day 1 ..... 3 1.1 USAMO 1996/1 ..... 3 1.2 USAMO 1996/2 ..... 4 1.3 USAMO 1996/3 2 Solutions to Day 2 ..... 6 2.1 USAMO 1996/4 ..... 6 2.2 USAMO 1996/5 ..... 7 2.3 USAMO 1996/6 ..... 8

§0 Problems

1. Prove that the average of the numbers $n \sin n^{\circ}$ for $n=2,4,6, \ldots, 180$ is $\cot 1^{\circ}$.
2. For any nonempty set $S$ of real numbers, let $\sigma(S)$ denote the sum of the elements of $S$. Given a set $A$ of $n$ positive integers, consider the collection of all distinct sums $\sigma(S)$ as $S$ ranges over the nonempty subsets of $A$. Prove that this collection of sums can be partitioned into $n$ classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2 .
3. Let $A B C$ be a triangle. Prove that there is a line $\ell$ (in the plane of triangle $A B C$ ) such that the intersection of the interior of triangle $A B C$ and the interior of its reflection $A^{\prime} B^{\prime} C^{\prime}$ in $\ell$ has area more than $\frac{2}{3}$ the area of triangle $A B C$.
4. An $n$-term sequence $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ in which each term is either 0 or 1 is called a binary sequence of length $n$. Let $a_{n}$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to $0,1,0$ in that order. Let $b_{n}$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to $0,0,1,1$ or $1,1,0,0$ in that order. Prove that $b_{n+1}=2 a_{n}$ for all positive integers $n$.
5. Let $A B C$ be a triangle, and $M$ an interior point such that $\angle M A B=10^{\circ}, \angle M B A=$ $20^{\circ}, \angle M A C=40^{\circ}$ and $\angle M C A=30^{\circ}$. Prove that the triangle is isosceles.
6. Determine with proof whether there is a subset $X \subseteq \mathbb{Z}$ with the following property: for any $n \in \mathbb{Z}$, there is exactly one solution to $a+2 b=n$, with $a, b \in X$.

§1 Solutions to Day 1

§1.1 USAMO 1996/1

Available online at https://aops.com/community/p353049.

Problem statement

Prove that the average of the numbers $n \sin n^{\circ}$ for $n=2,4,6, \ldots, 180$ is $\cot 1^{\circ}$.

Because

$$n\sin n^{\circ }+(180-n)\sin ({180}^{\circ }-n^{\circ })=180\sin n^{\circ }n\; \backslash sin\; n^\{\backslash circ\}+(180-n)\; \backslash sin\; \backslash left(180^\{\backslash circ\}-n^\{\backslash circ\}\backslash right)=180\; \backslash sin\; n^\{\backslash circ\}$$

So enough to show that

$$\sum _{n=0}^{89}\sin (2n{)}^{\circ }=\cot 1^{\circ }\backslash sum\_\{n=0\}^\{89\}\; \backslash sin\; (2\; n)^\{\backslash circ\}=\backslash cot\; 1^\{\backslash circ\}$$

Let $\zeta=\cos 2^{\circ}+i \sin 2^{\circ}$ be a primitive root. Then

Also,

So we're done.

§1.2 USAMO 1996/2

Available online at https://aops.com/community/p353051.

Problem statement

For any nonempty set $S$ of real numbers, let $\sigma(S)$ denote the sum of the elements of $S$. Given a set $A$ of $n$ positive integers, consider the collection of all distinct sums $\sigma(S)$ as $S$ ranges over the nonempty subsets of $A$. Prove that this collection of sums can be partitioned into $n$ classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.

By induction on $n$ with $n=1$ being easy. For the inductive step, assume

$$A=\{a_1>a_2>\cdots >a_n\}A=\backslash left\backslash \{a\_\{1\}>a\_\{2\}>\backslash cdots>a\_\{n\}\backslash right\backslash \}$$

Fix any index $k$ with the property that

$$a_k>\frac{\sigma (A)}{2^k}a\_\{k\}>\backslash frac\{\backslash sigma(A)\}\{2^\{k\}\}$$

(which must exist since $\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{k}}<1$ ). Then

• We make $k$ classes for the sums between $\frac{\sigma(A)}{2^{k}}$ and $\sigma(A)$; this handles every set which has any element in $\left{a_{1}, \ldots, a_{k}\right}$.
• We make $n-k$ classes via induction hypothesis on $\left{a_{k+1}, \ldots, a_{n}\right}$.

This solves the problem.

§1.3 USAMO 1996/3

Available online at https://aops.com/community/p353052.

Problem statement

Let $A B C$ be a triangle. Prove that there is a line $\ell$ (in the plane of triangle $A B C$ ) such that the intersection of the interior of triangle $A B C$ and the interior of its reflection $A^{\prime} B^{\prime} C^{\prime}$ in $\ell$ has area more than $\frac{2}{3}$ the area of triangle $A B C$.

All that's needed is: Claim - If $A B C$ is a triangle where $\frac{1}{2}<\frac{A B}{A C}<1$, then the $\angle A$ bisector works.

Proof. Let the $\angle A$-bisector meet $B C$ at $D$. The overlapped area is $2[A B D]$ and

$$\frac{[ABD]}{[ABC]}=\frac{BD}{BC}=\frac{AB}{AB+AC}\backslash frac\{[A\; B\; D]\}\{[A\; B\; C]\}=\backslash frac\{B\; D\}\{B\; C\}=\backslash frac\{A\; B\}\{A\; B+A\; C\}$$

by angle bisector theorem. In general, suppose $x<y<z$ are sides of a triangle. Then $\frac{1}{2}<\frac{y}{z}<1$ by triangle inequality as needed.

§2 Solutions to Day 2

§2.1 USAMO 1996/4

Available online at https://aops.com/community/p353058.

Problem statement

An $n$-term sequence $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ in which each term is either 0 or 1 is called a binary sequence of length $n$. Let $a_{n}$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to $0,1,0$ in that order. Let $b_{n}$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to $0,0,1,1$ or $1,1,0,0$ in that order. Prove that $b_{n+1}=2 a_{n}$ for all positive integers $n$.

Consider the map from sequences of the latter form to sequences of the first form by

$$(y_1,\dots ,y_{n+1})\mapsto (y_1+y_2,y_2+y_3,\dots ,y_n+y_{n+1})\mathrm{.}\backslash left(y\_\{1\},\; \backslash ldots,\; y\_\{n+1\}\backslash right)\; \backslash mapsto\backslash left(y\_\{1\}+y\_\{2\},\; y\_\{2\}+y\_\{3\},\; \backslash ldots,\; y\_\{n\}+y\_\{n+1\}\backslash right)\; .$$

It is 2-to-1. The end.

§2.2 USAMO 1996/5

Available online at https://aops.com/community/p2.

Problem statement

Let $A B C$ be a triangle, and $M$ an interior point such that $\angle M A B=10^{\circ}, \angle M B A=$ $20^{\circ}, \angle M A C=40^{\circ}$ and $\angle M C A=30^{\circ}$. Prove that the triangle is isosceles.

Let $\theta=\angle M B C<80^{\circ}$. By trig Ceva, we get

$$\frac{\sin {10}^{\circ }}{\sin {40}^{\circ }}\cdot \frac{\sin \theta }{\sin {20}^{\circ }}\cdot \frac{\sin {30}^{\circ }}{\sin ({80}^{\circ }-\theta )}=1\backslash frac\{\backslash sin\; 10^\{\backslash circ\}\}\{\backslash sin\; 40^\{\backslash circ\}\}\; \backslash cdot\; \backslash frac\{\backslash sin\; \backslash theta\}\{\backslash sin\; 20^\{\backslash circ\}\}\; \backslash cdot\; \backslash frac\{\backslash sin\; 30^\{\backslash circ\}\}\{\backslash sin\; \backslash left(80^\{\backslash circ\}-\backslash theta\backslash right)\}=1$$

This simplifies to

$$\sin \theta =4\sin ({80}^{\circ }-\theta )\sin {40}^{\circ }\cos {10}^{\circ }\backslash sin\; \backslash theta=4\; \backslash sin\; \backslash left(80^\{\backslash circ\}-\backslash theta\backslash right)\; \backslash sin\; 40^\{\backslash circ\}\; \backslash cos\; 10^\{\backslash circ\}$$

Claim - We have $\theta=60^{\circ}$. Proof. The left-hand side is increasing in $\theta$ and the right-hand side is decreasing in $\theta$, so at most one value of $\theta$ works. But we also have

as desired.

§2.3 USAMO 1996/6

Available online at https://aops.com/community/p353061.

Problem statement

Determine with proof whether there is a subset $X \subseteq \mathbb{Z}$ with the following property: for any $n \in \mathbb{Z}$, there is exactly one solution to $a+2 b=n$, with $a, b \in X$.

The idea is generating functions, but extra care is required since exponents will be in $\mathbb{Z}$ rather than in $\mathbb{Z}_{\geq 0}$.

However, consider formally the limit

$$f(x)=\prod _{k\ge 0}(1+x^{(-4{)}^k})f(x)=\backslash prod\_\{k\; \backslash geq\; 0\}\backslash left(1+x^\{(-4)^\{k\}\}\backslash right)$$

For size reasons, this indeed converges formally to a power series, in the sense that the coefficient of any $x^{k}$ is eventually zero or one for all partial sums.

We claim $X=\left{n:\left[x^{n}\right] f(x)=1\right}$ works. For a given $n$, we can truncate the sum at some large $N$ again for size reasons. For convenience assume $N$ is even. Now set

$$f_N(x)=\prod _{k=0}^N(1+x^{(-4{)}^k})f\_\{N\}(x)=\backslash prod\_\{k=0\}^\{N\}\backslash left(1+x^\{(-4)^\{k\}\}\backslash right)$$

Next, we compute

$$f_N(x)f_N\left(x^2\right)=\frac{(1+x)(1+x^2)\dots (1+x^{2^{2N+1}})}{x^{2+8+\cdots +2^{2N-1}}}=\cdots +x^{-2}+x^{-1}+1+x+x^2+\dots f\_\{N\}(x)\; f\_\{N\}\backslash left(x^\{2\}\backslash right)=\backslash frac\{(1+x)\backslash left(1+x^\{2\}\backslash right)\; \backslash ldots\backslash left(1+x^\{2^\{2\; N+1\}\}\backslash right)\}\{x^\{2+8+\backslash cdots+2^\{2\; N-1\}\}\}=\backslash cdots+x^\{-2\}+x^\{-1\}+1+x+x^\{2\}+\backslash ldots$$

as desired.