Difference between revisions of "2021 AIME I Problems/Problem 2"
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Let <math>G</math> be the intersection of <math>AD</math> and <math>FC</math>. | Let <math>G</math> be the intersection of <math>AD</math> and <math>FC</math>. | ||
From vertical angles, we know that <math>\angle FGA= \angle DGC</math>. Also, given that <math>ABCD</math> and <math>AFCE</math> are rectangles, we know that <math>\angle AFG= \angle CDG=90 ^{\circ}</math>. | From vertical angles, we know that <math>\angle FGA= \angle DGC</math>. Also, given that <math>ABCD</math> and <math>AFCE</math> are rectangles, we know that <math>\angle AFG= \angle CDG=90 ^{\circ}</math>. | ||
Revision as of 17:39, 11 March 2021
Problem
In the diagram below, 
is a rectangle with side lengths 
and 
, and 
is a rectangle with side lengths 
and 
as shown. The area of the shaded region common to the interiors of both rectangles is 
, where 
and 
are relatively prime positive integers. Find 
.
![[asy] pair A, B, C, D, E, F; A = (0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]](http://latex.artofproblemsolving.com/c/e/7/ce7ef019f55e9d0cf7364f8d93782a020489c947.png)
Solution (Similar Triangles)
Let 
be the intersection of 
and 
.
From vertical angles, we know that 
. Also, given that and 
are rectangles, we know that 
.
Therefore, by AA similarity, we know that triangles 
and 
are similar.
Let 
. Then, we have 
. By similar triangles, we know that 
and 
. We have 
.
Solving for 
, we have 
.
The area of the shaded region is just 
.
Thus, the answer is 
.
See also
| 2021 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
