2025 SSMO Team Round Problems/Problem 6

Problem

The rhombus $PQRS$ has side length $3$ . The point $X$ lies on segment $PR$ such that line $QX$ is perpendicular to line $PS$ . Given $QX=2$ , the area of $PQRS$ can be written as $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

$[asy] unitsize(1cm); import geometry; pair P,Q,R,S,T,X,Y; P=(0,0); Q=(3,2); R=(6,0); S=(3,-2); T=(3,0); X=(1.667,0); Y=extension(Q,X,P,S); draw(P--Q--R--S--cycle); draw(Q--X,red); draw(X--Y,red+dashed); draw(P--R,blue); draw(Q--S,blue); label("$P$",P,dir(180)); label("$Q$",Q,dir(90)); label("$R$",R,dir(0)); label("$S$",S,dir(270)); label("$X$",X,dir(-45)); label("$Y$",Y,dir(-135)); label("$C$",T,dir(-45)); draw(rightanglemark(Q,T,X,5)); draw(rightanglemark(P,Y,Q,5)); dot(P); dot(Q); dot(R); dot(S); dot(T); dot(X); dot(Y); [/asy]$

Let $C$ denote the center of the rhombus and let $Y$ be the intersection of $PS$ and $QX$ . Notice that $\triangle QCX \sim \triangle PYX \sim \triangle PCS \cong \triangle PCQ$ , and that the ratio of the side lengths of $\triangle QCX$ to those of $\triangle PCQ$ is $\tfrac{QX}{PQ} = \tfrac{2}{3}$ .

Let $PC = a$ and $QC = b$ . By similar triangles, we have $CX = \tfrac{2}{3}b$ . Then, Pythagoras on $\triangle QCX$ and $\triangle PCQ$ yields the system of equations \begin{align*} b^2 + (\tfrac{2}{3}b)^2 &= 4 \\ a^2+b^2 &= 9. \end{align*}Solving, we find that $(a,b) = (\tfrac{9}{\sqrt{13}}, \tfrac{6}{\sqrt{13}})$ . The area of the rhombus is $2ab = \tfrac{108}{13}$ , and we extract $108+13 = \boxed{121}$ .

~Sedro

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2025 SSMO Team Round Problems/Problem 6

Problem

Solution