2023 AIME II Problems/Problem 9
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.