2022 AIME I Problems/Problem 11
Contents
Problem
Let be a parallelogram with A circle tangent to sides and intersects diagonal at points and with as shown. Suppose that and Then the area of can be expressed in the form where and are positive integers, and is not divisible by the square of any prime. Find
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=1m3pqCgwLFE
Solution 1 (No trig)
Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let be our tangents from the circle to the parallelogram. By the secant power of a point, the power of . Then . Similarly, the power of and . We let and label the diagram accordingly.
Notice that because . Let be the center of the circle. Since and intersect and , respectively, at right angles, we have is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from to and to , and both are equal to . Since , . Since and . We can now use Pythagorean theorem on ; we have and .
We know that because is a parallelogram. Using Pythagorean theorem on , . Therefore, base . Thus the area of the parallelogram is the base times the height, which is and the answer is
~KingRavi
Solution 2
Let the circle tangent to at separately, denote that
Using POP, it is very clear that , let , using LOC in ,, similarly, use LOC in , getting that . We use the second equation to minus the first equation, getting that , we can get .
Now applying LOC in , getting , solving this equation to get , then , , the area is leads to
~bluesoul,HarveyZhang
Solution 3
Denote by the center of the circle. Denote by the radius of the circle. Denote by , , the points that the circle meets , , at, respectively.
Because the circle is tangent to , , , , , , .
Because , , , are collinear.
Following from the power of a point, . Hence, .
Following from the power of a point, . Hence, .
Denote . Because and are tangents to the circle, .
Because is a right trapezoid, . Hence, . This can be simplified as
In , by applying the law of cosines, we have
Because , we get . Plugging this into Equation (1), we get .
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 4
Let be the circle, let be the radius of , and let the points at which is tangent to , , and be , , and , respectively. Note that PoP on and with respect to yields and . We can compute the area of in two ways:
1. By the half-base-height formula, .
2. We can drop altitudes from the center of to , , and , which have lengths , , and . Thus, .
Equating the two expressions for and solving for yields .
Let . By the Parallelogram Law, . Solving for yields . Thus, , for a final answer of .
~ Leo.Euler
Solution 5
Let be the circle, let be the radius of , and let the points at which is tangent to , , and be , , and , respectively. PoP on and with respect to yields
Let
In
Area is
vladimir.shelomovskii@gmail.com, vvsss
Solution 6 (Short and Sweet)
Let be the center of the circle. Let points and be the tangent points of lines and respectively to the circle. By Power of a Point, . Similarly, . Notice that since quadrilateral is symmetrical. Let intersect at . Then, is similar to . Therefore, . Let the length of , then . Solving we get . Doing the Pythagorean theorem on triangles and for sides and respectively, we obtain the equation where denotes the radius of the circle. Solving, we get . Additionally, quadrilateral is symmetrical so . Let and extend a perpendicular foot from to and call it . Then, is right with , , and . Taking the difference of squares, we get . The area of is . Therefore, the answer is
Video Solution
https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s
~Steven Chen (www.professorchenedu.com)
Video Solution 2 (Mathematical Dexterity)
https://www.youtube.com/watch?v=1nDKQkr9NaU
Video Solution 3 by OmegaLearn
https://youtu.be/LpOegT0fKy8?t=740
~ pi_is_3.14
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.