Difference between revisions of "2019 AIME I Problems/Problem 15"
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==Problem 15== | ==Problem 15== | ||
− | Let <math>\overline{AB}</math> be a chord of a circle <math>\omega</math>, and let <math>P</math> be a point on the chord <math>\overline{AB}</math>. Circle <math>\omega_1</math> passes through <math>A</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circle <math>\omega_2</math> passes through <math>B</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>P</math> and <math>Q</math>. Line <math>PQ</math> intersects <math>\omega</math> at <math>X</math> and <math>Y</math>. Assume that <math>AP=5</math>, <math>PB=3</math>, <math>XY=11</math>, and <math>PQ = sqrt | + | Let <math>\overline{AB}</math> be a chord of a circle <math>\omega</math>, and let <math>P</math> be a point on the chord <math>\overline{AB}</math>. Circle <math>\omega_1</math> passes through <math>A</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circle <math>\omega_2</math> passes through <math>B</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>P</math> and <math>Q</math>. Line <math>PQ</math> intersects <math>\omega</math> at <math>X</math> and <math>Y</math>. Assume that <math>AP=5</math>, <math>PB=3</math>, <math>XY=11</math>, and <math>PQ = </math>\sqrt{\frac mn}<math>)</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
==Solution 1== | ==Solution 1== |
Revision as of 14:31, 30 July 2020
Problem 15
Let be a chord of a circle
, and let
be a point on the chord
. Circle
passes through
and
and is internally tangent to
. Circle
passes through
and
and is internally tangent to
. Circles
and
intersect at points
and
. Line
intersects
at
and
. Assume that
,
,
, and
\sqrt{\frac mn}
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let
and
be the centers of
and
, respectively. There is a homothety at
sending
to
that sends
to
and
to
, so
. Similarly,
, so
is a parallelogram. Moreover,
whence
is cyclic. However,
so
is an isosceles trapezoid. Since
,
, so
is the midpoint of
.
By Power of a Point, . Since
and
,
and the requested sum is
.
(Solution by TheUltimate123)
Solution 2
Let the tangents to at
and
intersect at
. Then, since
,
lies on the radical axis of
and
, which is
. It follows that
Let
denote the midpoint of
. By the Midpoint of Harmonic Bundles Lemma,
whence
. Like above,
. Since
, we establish that
, from which
, and the requested sum is
.
(Solution by TheUltimate123)
Solution 3
Firstly we need to notice that is the middle point of
. Assume the center of circle
are
, respectively. Then
are collinear and
are collinear. Link
. Notice that,
. As a result,
and
. So we have parallelogram
. So
Notice that,
and
divide
into two equal length pieces, So we have
. As a result,
lie on one circle. So
. Notice that
, we have
. As a result,
. So
is the middle point of
.
Back to our problem. Assume ,
and
. Then we have
, that is,
. Also,
. Solve these above, we have
. As a result, we have
. So, we have
. As a result, our answer is
.
Solution By BladeRunnerAUG (Fanyuchen20020715).
Solution 4
Note that the tangents to the circles at and
intersect at a point
on
by radical center. Then, since
and
, we have
so
is cyclic. But if
is the center of
, clearly
is cyclic with diameter
, so
is the midpoint of
. Then, by Power of a Point,
and it is given that
. Thus
so
and the answer is
.
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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