# USAMO 2003 Solution Notes Evan Chen《陳誼廷》 15 April 2024 This is a compilation of solutions for the 2003 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 2003/1, proposed by Titu Andreescu ..... 3 1.2 USAMO 2003/2 ..... 4 1.3 USAMO 2003/3 ..... 6 2 Solutions to Day 2 ..... 7 2.1 USAMO 2003/4, proposed by Titu Andreescu, Zuming Feng ..... 7 2.2 USAMO 2003/5, proposed by Zuming Feng, Titu Andreescu ..... 8 2.3 USAMO 2003/6 ..... 9 ## §0 Problems 1. Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^{n}$ all of whose digits are odd. 2. A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers. 3. Let $n$ be a positive integer. For every sequence of integers $$ A=\left(a_{0}, a_{1}, a_{2}, \ldots, a_{n}\right) $$ satisfying $0 \leq a_{i} \leq i$, for $i=0, \ldots, n$, we define another sequence $$ t(A)=\left(t\left(a_{0}\right), t\left(a_{1}\right), t\left(a_{2}\right), \ldots, t\left(a_{n}\right)\right) $$ by setting $t\left(a_{i}\right)$ to be the number of terms in the sequence $A$ that precede the term $a_{i}$ and are different from $a_{i}$. Show that, starting from any sequence $A$ as above, fewer than $n$ applications of the transformation $t$ lead to a sequence $B$ such that $t(B)=B$. 4. Let $A B C$ be a triangle. A circle passing through $A$ and $B$ intersects segments $A C$ and $B C$ at $D$ and $E$, respectively. Lines $A B$ and $D E$ intersect at $F$, while lines $B D$ and $C F$ intersect at $M$. Prove that $M F=M C$ if and only if $M B \cdot M D=M C^{2}$. 5. Let $a, b, c$ be positive real numbers. Prove that $$ \frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+c+a)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leq 8 $$ 6. At the vertices of a regular hexagon are written six nonnegative integers whose sum is $2003^{2003}$. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number 0 appears at all six vertices. ## §1 Solutions to Day 1 ## §1.1 USAMO 2003/1, proposed by Titu Andreescu Available online at https://aops.com/community/p336189. ## Problem statement Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^{n}$ all of whose digits are odd. This is immediate by induction on $n$. For $n=1$ we take 5 ; moving forward if $M$ is a working $n$-digit number then exactly one of $$ \begin{aligned} & N_{1}=10^{n}+M \\ & N_{3}=3 \cdot 10^{n}+M \\ & N_{5}=5 \cdot 10^{n}+M \\ & N_{7}=7 \cdot 10^{n}+M \\ & N_{9}=9 \cdot 10^{n}+M \end{aligned} $$ is divisible by $5^{n+1}$; as they are all divisible by $5^{n}$ and $N_{k} / 5^{n}$ are all distinct. ## §1.2 USAMO 2003/2 Available online at https://aops.com/community/p336193. ## Problem statement A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers. Suppose $A B$ is a side of a polygon in the dissection, lying on diagonal $X Y$, with $X, A$, $B, Y$ in that order. Then $$ A B=X Y-X A-Y B $$ In this way, we see that it actually just suffices to prove the result for a quadrilateral. We present two approaches to this end. 【 First approach (trig). Consider quadrilateral $A B C D$. There are twelve angles one can obtain using three of its four vertices, three at each vertex; denote this set of 12 angles by $S$ Note that: - The law of cosines implies $\cos \theta \in \mathbb{Q}$ for each $\theta \in S$. - Hence, $(\sin \theta)^{2} \in \mathbb{Q}$ for $\theta \in S$. (This is because $\sin \theta^{2}+\cos ^{2} \theta$.) We say two angles $\theta_{1}$ and $\theta_{2}$ are equivalent if $\frac{\sin \theta_{1}}{\sin \theta_{2}}$ This is the same as saying, when $\sin \theta_{1}$ and $\sin \theta_{2}$ are written in simplest radical form, the part under the square root is the same. Now we contend: Claim - The angles $\angle B A C, \angle C A D, \angle B A D$ are equivalent. Proof. Note that $$ \mathbb{Q} \ni \cos (\angle B A D)=\cos \angle B A C \cos \angle C A D-\sin \angle B A C \sin \angle C A D $$ so $\angle B A C$ and $\angle C A D$ are equivalent. Then $$ \sin (\angle B A D)=\sin \angle B A C \cos \angle C A D+\cos \angle B A C \sin \angle C A D $$ implies $\angle B A D$ is equivalent to those two. Claim - The angles $\angle B A D, \angle D B A, \angle A D B$ are equivalent. Proof. Law of sines on $\triangle B A D$. Iterating the argument implies that all angles are equivalent. Now, if $A B$ and $C D$ meet at $E$, the law of sines on $\triangle A E B$, etc. implies the result. I Second approach (barycentric coordinates). To do this, we apply barycentric coordinates. Consider quadrilateral $A B D C$ (note the changed order of vertices), with $A=(1,0,0), B=(0,1,0), C=(0,0,1)$. Let $D=(x, y, z)$, with $x+y+z=1$. By hypothesis, each of the numbers $$ \begin{aligned} -a^{2} y z+b^{2}(1-x) z+c^{2}(1-x) y & =A D^{2} \\ a^{2}(1-y) z+b^{2} z x+c^{2}(1-y) x & =B D^{2} \\ -a^{2}(1-z) y-b^{2}(1-z) x+c^{2} x y & =C D^{2} \end{aligned} $$ is rational. Let $W=a^{2} y z+b^{2} z x+c^{2} x y$. Then, $$ \begin{aligned} b^{2} z+c^{2} y & =A D^{2}+W \\ a^{2} z+c^{2} x & =B D^{2}+W \\ a^{2} y+b^{2} x & =C D^{2}+W \end{aligned} $$ This implies that $A D^{2}+B D^{2}+2 W-c^{2}=2 S_{C} z$ and cyclically (as usual $2 S_{C}=a^{2}+b^{2}-c^{2}$ ). If any of $S_{A}, S_{B}, S_{C}$ are zero, then we deduce $W$ is rational. Otherwise, we have that $$ 1=x+y+z=\sum_{\mathrm{cyc}} \frac{A D^{2}+B D^{2}+2 W-c^{2}}{2 S_{C}} $$ which implies that $W$ is rational, because it appears with coefficient $\frac{1}{S_{A}}+\frac{1}{S_{B}}+\frac{1}{S_{C}} \neq 0$ (since $S_{B C}+S_{C A}+S_{A B}$ is actually the area of $A B C$ ). Hence from the rationality of $W$, we deduce that $x$ is rational as long as $S_{A} \neq 0$, and similarly for the others. So at most one of $x, y, z$ is irrational, but since $x+y+z=1$ this implies they are all rational. Finally, if $P=\overline{A D} \cap \overline{B C}$ then $A P=\frac{1}{y+z} A D$, so $A P$ is rational too, completing the proof. Remark. After the reduction to quadrilateral, a third alternate approach goes by quoting Putnam 2018 A6, reproduced below: Four points are given in the plane, with no three collinear, such that the squares of the $\binom{4}{2}=6$ pairwise distances are all rational. Show that the ratio of the areas between any two of the $\binom{4}{3}=4$ triangles determined by these points is also rational. If $A B C D$ is the quadrilateral, the heights from $C$ and $D$ to $A B$ have rational ratio. Letting $P=A C \cap B D$, we see $A P / A B$ can be shown as rational via coordinates, as needed. ## §1.3 USAMO 2003/3 Available online at https://aops.com/community/p336202. ## Problem statement Let $n$ be a positive integer. For every sequence of integers $$ A=\left(a_{0}, a_{1}, a_{2}, \ldots, a_{n}\right) $$ satisfying $0 \leq a_{i} \leq i$, for $i=0, \ldots, n$, we define another sequence $$ t(A)=\left(t\left(a_{0}\right), t\left(a_{1}\right), t\left(a_{2}\right), \ldots, t\left(a_{n}\right)\right) $$ by setting $t\left(a_{i}\right)$ to be the number of terms in the sequence $A$ that precede the term $a_{i}$ and are different from $a_{i}$. Show that, starting from any sequence $A$ as above, fewer than $n$ applications of the transformation $t$ lead to a sequence $B$ such that $t(B)=B$. We go by strong induction on $n$ with the base cases $n=1$ and $n=2$ done by hand. Consider two cases: - If $a_{0}=0$ and $a_{1}=1$, then $1 \leq t\left(a_{i}\right) \leq i$ for $i \geq 1$; now apply induction to $$ \left(t\left(a_{1}\right)-1, t\left(a_{2}\right)-1, \ldots, t\left(a_{n}\right)-1\right) $$ - Otherwise, assume that $a_{0}=a_{1}=\cdots=a_{k-1}=0$ but $a_{k} \neq 0$, where $k \geq 2$. Assume $k