Difference between revisions of "2023 AIME II Problems/Problem 9"
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+ | Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at two points <math>P</math> and <math>Q,</math> and their common tangent line closer to <math>P</math> intersects <math>\omega_1</math> and <math>\omega_2</math> at points <math>A</math> and <math>B,</math> respectively. The line parallel to <math>AB</math> that passes through <math>P</math> intersects <math>\omega_1</math> and <math>\omega_2</math> for the second time at points <math>X</math> and <math>Y,</math> respectively. Suppose <math>PX=10,</math> <math>PY=14,</math> and <math>PQ=5.</math> Then the area of trapezoid <math>XABY</math> is <math>m\sqrt{n},</math> where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n.</math> | ||
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+ | ==Solution 1== | ||
Denote by <math>O_1</math> and <math>O_2</math> the centers of <math>\omega_1</math> and <math>\omega_2</math>, respectively. | Denote by <math>O_1</math> and <math>O_2</math> the centers of <math>\omega_1</math> and <math>\omega_2</math>, respectively. |
Revision as of 13:53, 17 February 2023
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
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,
.
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.