2021 AIME I Problems/Problem 2
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[hide]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
Solution 4 (Similar triangles and area)
Again, let the intersection of and be . By AA similarity, with a ratio. Define as . Because of similar triangles, . Using , the area of the parallelogram is . Using , the area of the parallelogram is . These equations are equal, so we can solve for and obtain . Thus, , so the area of the parallelogram is .
~mathboy100
Solution 5
The intersection of AD and FC = P. The intersection of AE and BC = K. Let's set AP to x. CK also has to be x because of the properties of a parallelogram. Then PD and BK must be 11-x because the sum of the segments has to be 11. We can easily solve for PC by the Pythagorean Theorem. DC^2 + PD^2 = PC^2. 9 + (11-x)^2 = PC^2. After 10 seconds of simplifying, we get that PC = sqrt(x^2-22x+30).
FC = 9, and FP + PC = 9. PC = sqrt(x^2-22x+30), so FP = 9 - sqrt(x^2-22x+30).
Now we can apply the Pythagorean Theorem to triangle AFP. AF^2 + FP^2 = AP^2. 49 + (9 - (sqrt(x^2-22x+30)))^2 = x^2. After simplifying (took me 2 minutes on the test), we get that x = 35/4. If you don't believe me, then plug it into WolframAlpha.
Now we have to solve for the area of APCK. We know that the height is 3 because the height of the parallelogram is the same as the height of the smaller rectangle.
The area of a parallelogram is base * height. The base is x (or 35/4), and the height is 3.
Multiplying, (35/4)*3 = 105/4. m+n = 105+4 = 109. ~ishanvannadil2008
Video Solution #1
https://youtu.be/M3DsERqhiDk?t=275
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 |
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