Difference between revisions of "2019 AIME I Problems/Problem 15"
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But if <math>O</math> is the center of <math>\omega</math>, clearly <math>ZAOB</math> is cyclic with diameter <math>ZO</math>, so <math>\angle ZQO = 90^{\circ} \implies Q</math> is the midpoint of <math>XY</math>. Then, by Power of a Point, <math>PY \cdot PX = PA \cdot PB = 15</math> and it is given that <math>PY+PX = 11</math>. Thus <math>PY, PX = \frac{11 \pm \sqrt{61}}{2}</math> so <math>PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}</math> and the answer is <math>61+4 = \boxed{065}</math>. | But if <math>O</math> is the center of <math>\omega</math>, clearly <math>ZAOB</math> is cyclic with diameter <math>ZO</math>, so <math>\angle ZQO = 90^{\circ} \implies Q</math> is the midpoint of <math>XY</math>. Then, by Power of a Point, <math>PY \cdot PX = PA \cdot PB = 15</math> and it is given that <math>PY+PX = 11</math>. Thus <math>PY, PX = \frac{11 \pm \sqrt{61}}{2}</math> so <math>PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}</math> and the answer is <math>61+4 = \boxed{065}</math>. | ||
+ | ==solution 5== | ||
+ | |||
+ | Connect <math>AQ,QB</math>, since <math>\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}</math>, so <math>\angle{AQP}=\frac{\angle{AO_1P}}{2}=\angle{BQP}=\frac{\angle{BO_2P}}{2}, \angle{AQB}=\angle{AOB}</math> then, so <math>A,O,Q,B</math> are concyclic | ||
+ | |||
+ | We let <math>\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}=2\alpha</math>, it is clear that <math>\angle{BQP}=\alpha, \angle{O_1AP}=90^{\circ}-\alpha</math>, which leads to the conclusion <math>OQ\bot XY</math> which tells <math>Q</math> is the midpoint of <math>XY</math> | ||
+ | |||
+ | Then it is clear, <math>XP\cdot PY=15, XP=\frac{11-\sqrt{61}}{2}, PQ=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}</math> , the answer is <math>\boxed{065}</math> | ||
+ | |||
+ | ~bluesoul | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=14|after=Last Problem}} | {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:46, 15 July 2022
Contents
Problem
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)
Note
One may solve for first using PoAP,
. Then, notice that
is rational but
is not, also
. The most likely explanation for this is that
is the midpoint of
, so that
and
. Then our answer is
. One can rigorously prove this using the methods above
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(EGMO 9.17),
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
divides
into two equal length pieces, So we have
. As a result,
lie on one circle. So
. Notice that since
, 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). Edited by bgn4493.
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
.
solution 5
Connect , since
, so
then, so
are concyclic
We let , it is clear that
, which leads to the conclusion
which tells
is the midpoint of
Then it is clear, , the answer is
~bluesoul
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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