Difference between revisions of "2019 AIME I Problems/Problem 15"
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− | <math>PX \cdot PY=AP \cdot PB=5 \cdot 3=15</math> by power of a point. Also, <math>PX+PY=XY=11</math>, so <math>PX</math> and <math>PY</math> are solutions to the quadratic <math>x^2-11x+15=0</math> so PX and PY is <math>\frac{11\pm\sqrt{61}}{2}</math> in some order. Now, because we want <math>PQ^2</math> and it is known to be rational, we can guess that <math>PQ</math> is irrational or the problem would simply ask for <math>PQ</math>. <math>PQ=QX-PX</math>, and chances are low that <math>QX</math> is some number with a square root plus or minus <math>\frac{\sqrt{61}}{2}</math> to cancel out the <math>\frac{\sqrt{61}}{2}</math> in <math>PX</math>, so one can see that <math>PQ^2</math> is most likely to be <math>(\frac{\sqrt{61}}{2})^2=\frac{61}{4}</math>, and our answer is <math>61+4=\boxed{65}</math> | + | <math>PX \cdot PY=AP \cdot PB=5 \cdot 3=15</math> by power of a point. Also, <math>PX+PY=XY=11</math>, so <math>PX</math> and <math>PY</math> are solutions to the quadratic <math>x^2-11x+15=0</math> so <math>PX</math> and <math>PY</math> is <math>\frac{11\pm\sqrt{61}}{2}</math> in some order. Now, because we want <math>PQ^2</math> and it is known to be rational, we can guess that <math>PQ</math> is irrational or the problem would simply ask for <math>PQ</math>. <math>PQ=QX-PX</math>, and chances are low that <math>QX</math> is some number with a square root plus or minus <math>\frac{\sqrt{61}}{2}</math> to cancel out the <math>\frac{\sqrt{61}}{2}</math> in <math>PX</math>, so one can see that <math>PQ^2</math> is most likely to be <math>(\frac{\sqrt{61}}{2})^2=\frac{61}{4}</math>, and our answer is <math>61+4=\boxed{65}</math> |
Note : If our answer is correct, then <math>QX=\frac{11}{2}</math>, which made Q the midpoint of XY, a feature that occurs often in AIME problems, so that again made our answer probable, and even if it's wrong, it's still the same as leaving it blank because it's a AIME problem. ~ddk001 | Note : If our answer is correct, then <math>QX=\frac{11}{2}</math>, which made Q the midpoint of XY, a feature that occurs often in AIME problems, so that again made our answer probable, and even if it's wrong, it's still the same as leaving it blank because it's a AIME problem. ~ddk001 |
Revision as of 15:08, 4 November 2023
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 axis theorem. Since
and
, we have
so
is cyclic.
But if is the center of
, clearly
is cyclic with diameter
, so
implies that
is the midpoint of
. Then, by power of point
,
whereas it is given that
. Thus
so
, i.e.
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
Solution 6(calculationless)
by power of a point. Also,
, so
and
are solutions to the quadratic
so
and
is
in some order. Now, because we want
and it is known to be rational, we can guess that
is irrational or the problem would simply ask for
.
, and chances are low that
is some number with a square root plus or minus
to cancel out the
in
, so one can see that
is most likely to be
, and our answer is
Note : If our answer is correct, then , which made Q the midpoint of XY, a feature that occurs often in AIME problems, so that again made our answer probable, and even if it's wrong, it's still the same as leaving it blank because it's a AIME problem. ~ddk001
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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