Difference between revisions of "2019 AIME I Problems/Problem 13"
(Solution is incomplete. I just need to add the part about using similar triangles to find GF and BG) |
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==Solution== | ==Solution== | ||
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+ | Define <math>\omega_1</math> to be the circumcircle of <math>\triangle ACD</math> and <math>\omega_2</math> to be the circumcircle of <math>\triangle EBC</math>. | ||
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+ | Because of exterior angles, | ||
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+ | <math>\angle ACB = \angle CBE - \angle CAD</math> | ||
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+ | But <math>\angle CBE = \angle CFE</math> because <math>CBFE</math> is cyclic. In addition, <math>\angle CAD = \angle CFD</math> because <math>CAFD</math> is cyclic. Therefore, <math>\angle ACB = \angle CFE - \angle CFD</math>. But <math>\angle CFE - \angle CFD = \angle DFE</math>, so <math>\angle ACB = \angle DFE</math>. Using Law of Cosines on <math>\triangle ABC</math>, we can figure out that <math>\cos(\angle ACB) = \frac{3}{4}</math>. Since <math>\angle ACB = \angle DFE</math>, <math>\cos(\angle DFE) = \frac{3}{4}</math>. We are given that <math>DF = 2</math> and <math>FE = 7</math>, so we can use Law of Cosines on <math>\triangle DEF</math> to find that <math>DE = 4\sqrt{2}</math>. | ||
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+ | Let <math>G</math> be the intersection of segment <math>\overline{AE}</math> and <math>\overline{CF}</math>. Using Power of a Point with respect to <math>G</math> within <math>\omega_1</math>, we find that <math>AG \cdot GD = CG \cdot GF</math>. We can also apply Power of a Point with respect to <math>G</math> within <math>\omega_2</math> to find that <math>CG \cdot GF = BG \cdot GE</math>. Therefore, <math>AG \cdot GD = BG \cdot GE</math>. | ||
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+ | <math>AG \cdot GD = BG \cdot GE</math> | ||
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+ | <math>(AB + BG) \cdot GD = BG \cdot (GD + DE)</math> | ||
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+ | <math>AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE</math> | ||
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+ | <math>AB \cdot GD = BG \cdot DE</math> | ||
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+ | <math>4 \cdot GD = BG \cdot 4\sqrt{2}</math> | ||
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+ | <math>\frac{GD}{BG} = \sqrt{2}</math> | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=12|num-a=14}} | {{AIME box|year=2019|n=I|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 05:35, 15 March 2019
Problem 13
Triangle has side lengths
,
, and
. Points
and
are on ray
with
. The point
is a point of intersection of the circumcircles of
and
satisfying
and
. Then
can be expressed as
, where
,
,
, and
are positive integers such that
and
are relatively prime, and
is not divisible by the square of any prime. Find
.
Solution
Define to be the circumcircle of
and
to be the circumcircle of
.
Because of exterior angles,
But because
is cyclic. In addition,
because
is cyclic. Therefore,
. But
, so
. Using Law of Cosines on
, we can figure out that
. Since
,
. We are given that
and
, so we can use Law of Cosines on
to find that
.
Let be the intersection of segment
and
. Using Power of a Point with respect to
within
, we find that
. We can also apply Power of a Point with respect to
within
to find that
. Therefore,
.
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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