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Difference between revisions of "2021 CIME I Problems/Problem 1"

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== Solution ==
 
== Solution ==
−
<math>\boxed{123}</math>
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 +
<asy>
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import geometry;
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point B = (0,0);
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point C = (16,0);
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point A = (0,16);
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point D = (16,16);
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point P = A * 1/3;
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point Q = C * 38/85 + D * 47/85;
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// Square ABCD
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draw(A--B--C--D--A);
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dot(A);
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label("A",A,NW);
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dot(B);
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label("B",B,SW);
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dot(C);
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label("C",C,SE);
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dot(D);
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label("D",D,NE);
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// Segment PQ
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draw(P--Q);
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dot(P);
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label("P",P,W);
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dot(Q);
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label("Q",Q,E);
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</asy>
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 +
From the problem, we know that  <math>[ABCD]=[APQD]+[BPQC]=20+21=41</math>. Thus, the side length of the square is <math>\sqrt{41}</math>. Furthermore, because <math>\tfrac{AP}{BP}=2</math>, <math>AP = \tfrac{2}{3}\sqrt{41}</math>. Because <math>\overline{DQ} \parallel \overline{AP}</math>, <math>APQD</math> is a trapezoid. Thus, if we let <math>DQ=x</math>, the area of <math>APQD</math> is <math>\frac{x+\tfrac{2\sqrt{41}}{3}}{2}*\sqrt{41}</math>. Equating this to the given area of <math>20</math>, we can now solve for <math>x</math>:
 +
\begin{align*}
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\frac{x+\tfrac{2\sqrt{41}}{3}}{2}*\sqrt{41} &= 20 \\
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\frac{\sqrt{41}x}{2}+\frac{\sqrt{41}}{3}*\sqrt{41} &= 20 \\
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3\sqrt{41}x+82 &= 120 \\
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x &= \frac{38}{3\sqrt{41}} = \frac{38\sqrt{41}}{123}
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\end{align*}
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Because <math>CQ=\sqrt{41}-x</math>, we can now find a value for <math>\frac{DQ}{CQ}</math>:
 +
\begin{align*}
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\frac{DQ}{CQ} &= \frac{x}{\sqrt{41}-x} \\
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&= \frac{\tfrac{38\sqrt{41}}{123}}{\sqrt{41}-\tfrac{38\sqrt{41}}{123}} \\
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&= \frac{38}{123-38} \\
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&= \frac{38}{85}
 +
\end{align*}
 +
 
 +
Thus, our answer is <math>38+85=\boxed{123}</math>.
  
 
== See also ==
 
== See also ==
 
{{CIME box|year=2021|n=I|before=First Problem|num-a=2}}
 
{{CIME box|year=2021|n=I|before=First Problem|num-a=2}}

Latest revision as of 18:26, 24 July 2024

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 $38+85=\boxed{123}$.

See also

2021 CIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All CIME Problems and Solutions
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