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 04:35, 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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