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
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
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
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 |
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