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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 $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.

Solution

Define $\omega_1$ to be the circumcircle of $\triangle ACD$ and $\omega_2$ to be the circumcircle of $\triangle EBC$.

Because of exterior angles,

$\angle ACB = \angle CBE - \angle CAD$

But $\angle CBE = \angle CFE$ because $CBFE$ is cyclic. In addition, $\angle CAD = \angle CFD$ because $CAFD$ is cyclic. Therefore, $\angle ACB = \angle CFE - \angle CFD$. But $\angle CFE - \angle CFD = \angle DFE$, so $\angle ACB = \angle DFE$. Using Law of Cosines on $\triangle ABC$, we can figure out that $\cos(\angle ACB) = \frac{3}{4}$. Since $\angle ACB = \angle DFE$, $\cos(\angle DFE) = \frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\triangle DEF$ to find that $DE = 4\sqrt{2}$.

Let $G$ be the intersection of segment $\overline{AE}$ and $\overline{CF}$. Using Power of a Point with respect to $G$ within $\omega_1$, we find that $AG \cdot GD = CG \cdot GF$. We can also apply Power of a Point with respect to $G$ within $\omega_2$ to find that $CG \cdot GF = BG \cdot GE$. Therefore, $AG \cdot GD = BG \cdot GE$.

$AG \cdot GD = BG \cdot GE$

$(AB + BG) \cdot GD = BG \cdot (GD + DE)$

$AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE$

$AB \cdot GD = BG \cdot DE$

$4 \cdot GD = BG \cdot 4\sqrt{2}$

$\frac{GD}{BG} = \sqrt{2}$

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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