Difference between revisions of "2021 AIME I Problems/Problem 2"
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Solving for <math>x</math>, we have <math>x=\frac{35}{4}</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>. | 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>. | + | Thus, the answer is <math>105+4=\framebox{109}</math>. ~yuanyuanC |
==See also== | ==See also== |
Revision as of 17:58, 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 (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 . ~yuanyuanC
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.