Difference between revisions of "2019 AIME I Problems/Problem 15"
m (→Solution 5 (Easy)) |
(Tag: Undo) |
||
Line 62: | Line 62: | ||
so <math>ZAQB</math> is cyclic. 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>. | so <math>ZAQB</math> is cyclic. 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 ( | + | ==Solution 5 (Easy)== |
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>. | 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>. | ||
Revision as of 14:50, 2 June 2020
Contents
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 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
.
Solution 5 (Easy)
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
.
-Albert Einstein
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.