Difference between revisions of "2021 AIME I Problems/Problem 2"
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==Solution== | ==Solution== | ||
+ | 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>. | ||
+ | Therefore, by AA similarity, we know that triangles <math>AFG</math> and <math>CDG</math> are similar. | ||
+ | Let <math>AG=x</math>. Then, we have <math>DG=11-x</math>. By similar triangles, we know that <math>FG=\frac{7}{3}(11-x)</math> and <math>CG=\frac{3}{7}x</math>. We have <math>\frac{7}{3}(11-x)+\frac{3}{7}x=FC=9</math>. | ||
+ | Solving for <math>x</math>, we have <math>x=\frac{35}{4}</math>. | ||
+ | The area of the shaded region is just <math>3\cdot \frac{35}{4}=\frac{105}{4}</math>. | ||
+ | Thus, the answer is <math>105+4=\framebox{109}</math>. | ||
==See also== | ==See also== |
Revision as of 17:37, 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 .
Solution
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.