Difference between revisions of "2019 AIME I Problems/Problem 15"
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+ | ==Solution 6(calculationless)== | ||
+ | <math>PX\cdotPY=AP\cdotPB=5\cdot3=15</math> by power of a point. Also, <math>PX+PY=XY=11</math>, so PX and PY are solutions to the quadratic <math>x^2-11x+15=0</math> so PX and PY is <math>\frac{11\mp\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. | ||
==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 13:52, 4 November 2023
Contents
[hide]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)
$PX\cdotPY=AP\cdotPB=5\cdot3=15$ (Error compiling LaTeX. Unknown error_msg) by power of a point. Also, , so PX and PY are solutions to the quadratic
so PX and PY is $\frac{11\mp\sqrt{61}{2}}$ (Error compiling LaTeX. Unknown error_msg) 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.
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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