2023 AIME II Problems/Problem 9
Contents
Problem
Circles and intersect at two points and and their common tangent line closer to intersects and at points and respectively. The line parallel to that passes through intersects and for the second time at points and respectively. Suppose and Then the area of trapezoid is where and are positive integers and is not divisible by the square of any prime. Find
Video Solution & More by MegaMath
https://www.youtube.com/watch?v=x-5VYR1Dfw4
Solution 1
Denote by and the centers of and , respectively. Let and intersect at point . Let and intersect at point .
Because is tangent to circle , . Because , . Because and are on , is the perpendicular bisector of . Thus, .
Analogously, we can show that .
Thus, . Because , , , , is a rectangle. Hence, .
Let and meet at point . Thus, is the midpoint of . Thus, . This is the case because is the radical axis of the two circles, and the powers with respect to each circle must be equal.
In , for the tangent and the secant , following from the power of a point, we have . By solving this equation, we get .
We notice that is a right trapezoid. Hence,
Therefore,
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~Mathkiddie
Solution 2
Notice that line is the radical axis of circles and . By the radical axis theorem, we know that the tangents of any point on line to circles and are equal. Therefore, line must pass through the midpoint of , call this point M. In addition, we know that by circle properties and midpoint definition.
Then, by Power of Point,
Call the intersection point of line and be C, and the intersection point of line and be D. is a rectangle with segment drawn through it so that , , and . Dropping the altitude from to , we get that the height of trapezoid is . Therefore the area of trapezoid is
Giving us an answer of .
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=RUv6qNY_agI
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.