Difference between revisions of "2019 AIME I Problems/Problem 15"
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Back to our problem. Assume <math>XP=x</math>, <math>PY=y</math> and <math>x<y</math>. Then we have <math>AP\cdot PB=XP\cdot PY</math>, that is, <math>xy=15</math>. Also, <math>XP+PY=x+y=XY=11</math>. Solve these above, we have <math>x=\frac{11-\sqrt{61}}{2}=XP</math>. As a result, we hav e <math>PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}</math>. So, we have <math>PQ^2=\frac{61}{4}</math>. As a result, our answer is <math>m+n=61+4=\boxed{065}</math>. | Back to our problem. Assume <math>XP=x</math>, <math>PY=y</math> and <math>x<y</math>. Then we have <math>AP\cdot PB=XP\cdot PY</math>, that is, <math>xy=15</math>. Also, <math>XP+PY=x+y=XY=11</math>. Solve these above, we have <math>x=\frac{11-\sqrt{61}}{2}=XP</math>. As a result, we hav e <math>PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}</math>. So, we have <math>PQ^2=\frac{61}{4}</math>. As a result, our answer is <math>m+n=61+4=\boxed{065}</math>. | ||
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Solution By BladeRunnerAUG (Fanyuchen20020715). | Solution By BladeRunnerAUG (Fanyuchen20020715). | ||
+ | ==Solution 4 (Lazy)== | ||
+ | First we solve for <math>PX</math> with PoAP, <math>PX = \frac{11}{2} - \frac{\sqrt{61}}{2}</math>. Notice that <math>PQ^2</math> is rational but <math>PX^2</math> is not, also <math>PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}</math>. The most likely explanation for this is that <math>Q</math> is the midpoint of <math>XY</math>, so that <math>XQ = \frac{11}{2}</math> and <math>PQ=\frac{\sqrt{61}}{2}</math>. Then our answer is <math>m+n=61+4=\boxed{065}</math>. | ||
==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 11:31, 17 March 2019
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).
Solution 4 (Lazy)
First we solve for with PoAP,
. 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
.
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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