Difference between revisions of "2023 AIME II Problems/Problem 9"
(Created page with "==Solution== Denote by <math>O_1</math> and <math>O_2</math> the centers of <math>\omega_1</math> and <math>\omega_2</math>, respectively. Let <math>XY</math> and <math>AO_1<...") |
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<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | + | [XABY] & = \frac{1}{2} \left( AB + XY \right) AC \\ | |
& = \frac{1}{2} \left( 12 + 24 \right) \sqrt{15} \\ | & = \frac{1}{2} \left( 12 + 24 \right) \sqrt{15} \\ | ||
& = 18 \sqrt{15}. | & = 18 \sqrt{15}. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2023|num-b=8|num-a=10|n=II}} | ||
+ | {{MAA Notice}} |
Revision as of 19:02, 16 February 2023
Solution
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, .
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)
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.