2023 AIME II Problems/Problem 9
Contents
[hide]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 .
Solution 3
Refer to Solution 1.
We let and the extension of
to the circle
as
By Power of a Point on point
of circle
we find
We have diameter
Therefore the radius of
is
Similarly repeating this procedure on we find the radius of
is
Next we solve for in two ways. Let the perpendicular from
to
intersect at
we have
We also have
Therefore since
is right, we have
For our second way, we let the midpoint of be
Note that
forms the right triangles
and
both of which share an leg of
or
Using Pythag we can solve for
We let
to slightly simplify the equation,
Thus the solutions are
Checking bounds
is the only valid solution, which means
Finally to find the area of
we have the bases
and
and the height
therefore
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 |
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