Difference between revisions of "2021 AIME I Problems/Problem 2"
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==See also== | ==See also== |
Revision as of 01:36, 12 March 2021
Contents
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 1 (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
Solution 2 (Coordinate Geometry Bash)
Suppose It follows that
Since is a rectangle, we have and The equation of the circle with center and radius is and the equation of the circle with center and radius is
We now have a system of two equations with two variables. Expanding and rearranging respectively give Subtracting from we get Simplifying and rearranging produce Substituting into gives which is a quadratic of We clear fractions by multiplying both sides by then solve by factoring: Since is in Quadrant IV, we have It follows that the equation of is
Let be the intersection of and and be the intersection of and Since is the -intercept of we obtain
By symmetry, quadrilateral is a parallelogram. Its area is and the requested sum is
~MRENTHUSIASM
Solution 3 (Pythagorean Theorem)
Let the intersection of and be , and let , so .
By the Pythagorean theorem, , so , and thus .
By the Pythagorean theorem again, :
Solving, we get , so the area of the parallelogram is , and .
~JulianaL25
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY&t=289s
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.