2021 CIME I Problems/Problem 1

Problem

Let $ABCD$ be a square. Points $P$ and $Q$ are on sides $AB$ and $CD,$ respectively $,$ such that the areas of quadrilaterals $APQD$ and $BPQC$ are $20$ and $21,$ respectively. Given that $\tfrac{AP}{BP}=2,$ then $\tfrac{DQ}{CQ}=\tfrac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. Find $a+b$ .

Solution

$[asy] import geometry; point B = (0,0); point C = (16,0); point A = (0,16); point D = (16,16); point P = A * 1/3; point Q = C * 38/85 + D * 47/85; // Square ABCD draw(A--B--C--D--A); dot(A); label("A",A,NW); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(D); label("D",D,NE); // Segment PQ draw(P--Q); dot(P); label("P",P,W); dot(Q); label("Q",Q,E); [/asy]$

From the problem, we know that $[ABCD]=[APQD]+[BPQC]=20+21=41$ . Thus, the side length of the square is $\sqrt{41}$ . Furthermore, because $\tfrac{AP}{BP}=2$ , $AP = \tfrac{2}{3}\sqrt{41}$ . Because $\overline{DQ} \parallel \overline{AP}$ , $APQD$ is a trapezoid. Thus, if we let $DQ=x$ , the area of $APQD$ is $\frac{x+\tfrac{2\sqrt{41}}{3}}{2}*\sqrt{41}$ . Equating this to the given area of $20$ , we can now solve for $x$ : \begin{align*} \frac{x+\tfrac{2\sqrt{41}}{3}}{2}*\sqrt{41} &= 20 \\ \frac{\sqrt{41}x}{2}+\frac{\sqrt{41}}{3}*\sqrt{41} &= 20 \\ 3\sqrt{41}x+82 &= 120 \\ x &= \frac{38}{3\sqrt{41}} = \frac{38\sqrt{41}}{123} \end{align*} Because $CQ=\sqrt{41}-x$ , we can now find a value for $\frac{DQ}{CQ}$ : \begin{align*} \frac{DQ}{CQ} &= \frac{x}{\sqrt{41}-x} \\ &= \frac{\tfrac{38\sqrt{41}}{123}}{\sqrt{41}-\tfrac{38\sqrt{41}}{123}} \\ &= \frac{38}{123-38} \\ &= \frac{38}{85} \end{align*}

Thus, our answer is $38+85=\boxed{123}$ .

2021 CIME I (Problems • Answer Key • Resources)
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2021 CIME I Problems/Problem 1

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