Difference between revisions of "2021 CIME I Problems/Problem 1"
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− | <math>\boxed{123}</math> | + | 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*} | ||
+ | \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 <math>CQ=\sqrt{41}-x</math>, we can now find a value for <math>\frac{DQ}{CQ}</math>: | ||
+ | \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 <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 be a square. Points and are on sides and respectively such that the areas of quadrilaterals and are and respectively. Given that then where and are relatively prime positive integers. Find .
Solution
From the problem, we know that . Thus, the side length of the square is . Furthermore, because , . Because , is a trapezoid. Thus, if we let , the area of is . Equating this to the given area of , we can now solve for : \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 , we can now find a value for : \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 .
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 |