Difference between revisions of "2019 AIME I Problems/Problem 15"
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Let <math>Q'</math> denote the midpoint of <math>\overline{XY}</math>. By the Midpoint of Harmonic Bundles Lemma, <cmath>RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,</cmath> | Let <math>Q'</math> denote the midpoint of <math>\overline{XY}</math>. By the Midpoint of Harmonic Bundles Lemma, <cmath>RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,</cmath> | ||
whence <math>Q=Q'</math>. Like above, <math>XP=\frac{11-\sqrt{61}}2</math>. Since <math>XQ=\frac{11}2</math>, we establish that <math>PQ=\frac{\sqrt{61}}2</math>, from which <math>PQ^2=\frac{61}4</math>, and the requested sum is <math>61+4=\boxed{065}</math>. | whence <math>Q=Q'</math>. Like above, <math>XP=\frac{11-\sqrt{61}}2</math>. Since <math>XQ=\frac{11}2</math>, we establish that <math>PQ=\frac{\sqrt{61}}2</math>, from which <math>PQ^2=\frac{61}4</math>, and the requested sum is <math>61+4=\boxed{065}</math>. | ||
+ | |||
+ | (Solution by TheUltimate123) | ||
==Solution 3== | ==Solution 3== |
Revision as of 14:07, 16 March 2019
Contents
[hide]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
, 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 hav e
. So, we have
. As a result, our answer is
.
Solution By BladeRunnerAUG (Fanyuchen20020715).
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.