{"year": "2003", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^{n}$ all of whose digits are odd.", "solution": " 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.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2003-notes.jsonl"}} {"year": "2003", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "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.", "solution": " 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. 【 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. $$ \\mathbb{Q} \\ni \\cos (\\angle B A D)=\\cos \\angle B A C \\cos \\angle C A D-\\sin \\angle B A C \\sin \\angle C A D $$ 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. 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.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2003-notes.jsonl"}} {"year": "2003", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "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.", "solution": " 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. 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.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2003-notes.jsonl"}} {"year": "2003", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "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 $t(B)=B$.", "solution": " 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