2021 AIME I Problems/Problem 2
Contents
- 1 Problem
- 2 Solution 1 (Similar Triangles)
- 3 Solution 2 (Similar Triangles)
- 4 Solution 3 (Pythagorean Theorem)
- 5 Solution 4 (Pythagorean Theorem)
- 6 Solution 5 (Coordinate Geometry)
- 7 Solution 6 (Trigonometry)
- 8 Video Solution by Punxsutawney Phil
- 9 Video Solution
- 10 Video Solution by Steven Chen (in Chinese)
- 11 Video Solution
- 12 Video Solution by Power of Logic
- 13 See Also
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, because we are given that and are rectangles, we know that . Therefore, by AA similarity, we know that .
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 (Similar Triangles)
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 .
Finally, the answer is .
~mathboy100
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 (Pythagorean Theorem)
Let , and . Also let .
also has to be by parallelogram properties. Then and must be because the sum of the segments has to be .
We can easily solve for by the Pythagorean Theorem: It follows shortly that .
Also, , and . We can then say that , so .
Now we can apply the Pythagorean Theorem to .
This simplifies (not-as-shortly) to . Now we have to solve for the area of . We know that the height is because the height of the parallelogram is the same as the height of the smaller rectangle.
From the area of a parallelogram (we know that the base is and the height is ), it is clear that the area is , giving an answer of .
~ishanvannadil2008 (Solution Sketch)
~Tuatara (Rephrasing and )
Solution 5 (Coordinate Geometry)
Suppose It follows that and
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 obtain 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 get
By symmetry, quadrilateral is a parallelogram. Its area is from which the requested sum is
~MRENTHUSIASM
Solution 6 (Trigonometry)
Let the intersection of and be . It is useful to find , because and . From there, subtracting the areas of the two triangles from the larger rectangle, we get Area = .
let . Let . Note, .
. The answer is .
~twotothetenthis1024
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY&t=289s
Video Solution
https://youtu.be/M3DsERqhiDk?t=275
Video Solution by Steven Chen (in Chinese)
Video Solution
https://www.youtube.com/watch?v=BinfKrc5bWo
Video Solution by Power of Logic
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.