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> | 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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+ | ==Video Solution & More by MegaMath== | ||
+ | https://www.youtube.com/watch?v=x-5VYR1Dfw4 | ||
==Solution 1== | ==Solution 1== | ||
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Let <math>QP</math> and <math>AB</math> meet at point <math>M</math>. | Let <math>QP</math> and <math>AB</math> meet at point <math>M</math>. | ||
Thus, <math>M</math> is the midpoint of <math>AB</math>. | Thus, <math>M</math> is the midpoint of <math>AB</math>. | ||
− | Thus, <math>AM = \frac{AB}{2} = 6</math> | + | Thus, <math>AM = \frac{AB}{2} = 6</math>. This is the case because <math>PQ</math> is the radical axis of the two circles, and the powers with respect to each circle must be equal. |
In <math>\omega_1</math>, for the tangent <math>MA</math> and the secant <math>MPQ</math>, following from the power of a point, we have <math>MA^2 = MP \cdot MQ</math>. | In <math>\omega_1</math>, for the tangent <math>MA</math> and the secant <math>MPQ</math>, following from the power of a point, we have <math>MA^2 = MP \cdot MQ</math>. | ||
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Therefore, the answer is <math>18 + 15 = \boxed{\textbf{(033)}}</math>. | Therefore, the answer is <math>18 + 15 = \boxed{\textbf{(033)}}</math>. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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==Solution 2== | ==Solution 2== |
Latest revision as of 16:14, 7 June 2024
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)
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.