2021 CIME I Problems/Problem 1
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
![[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]](http://latex.artofproblemsolving.com/4/1/e/41e67c0930c74ea344606d3c8ea9f42444338435.png)
From the problem, we know that ![$[ABCD]=[APQD]+[BPQC]=20+21=41$](http://latex.artofproblemsolving.com/0/8/9/089e05f7fcb8c9233b12f876225f7313346a9cd4.png)
. 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 | ||
