Difference between revisions of "2018 AIME II Problems/Problem 14"
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<cmath> -1 = (A,O;Y,X) \stackrel{Z}{=} (A,M;P,B) \stackrel{Z}{=} (A,N;Q,C).</cmath>Therefore, <math>\frac{AP \cdot MB}{MP \cdot AB} = 1 \implies \frac{3 \cdot MB}{MP \cdot 7} = 1</math>. Since <math>MB+MP=4</math> we know that <math>MB = \frac{6}{5}</math> and <math>MB = \frac{14}{5}</math>. Therefore, <math>AN = AM = \frac{21}{5}</math> and <math>NC = 8 - \frac{21}{5} = \frac{19}{5}</math>. Since <math>(A,N;Q,C) = -1</math>, we also have <math>\frac{AQ \cdot NC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1</math>. Solving for <math>AQ</math>, we obtain <math>AQ = \frac{168}{59} \implies m+n = \boxed{227}</math>. | <cmath> -1 = (A,O;Y,X) \stackrel{Z}{=} (A,M;P,B) \stackrel{Z}{=} (A,N;Q,C).</cmath>Therefore, <math>\frac{AP \cdot MB}{MP \cdot AB} = 1 \implies \frac{3 \cdot MB}{MP \cdot 7} = 1</math>. Since <math>MB+MP=4</math> we know that <math>MB = \frac{6}{5}</math> and <math>MB = \frac{14}{5}</math>. Therefore, <math>AN = AM = \frac{21}{5}</math> and <math>NC = 8 - \frac{21}{5} = \frac{19}{5}</math>. Since <math>(A,N;Q,C) = -1</math>, we also have <math>\frac{AQ \cdot NC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1</math>. Solving for <math>AQ</math>, we obtain <math>AQ = \frac{168}{59} \implies m+n = \boxed{227}</math>. | ||
-Vfire | -Vfire | ||
+ | |||
+ | ==Solution 3 (Combination of Law of Sine and Law of Cosine)== | ||
+ | Let the center of the incircle of <math>\triangle ABC</math> be <math>O</math>. Link <math>OY</math> and <math>OX</math>. Then we have <math>\angle OYP=\angle OXB=90^{\circ}</math> | ||
+ | |||
+ | <math>\because</math> <math>OY=OX</math> | ||
+ | |||
+ | <math>\therefore</math> <math>\angle OYX=\angle OXY</math> | ||
+ | |||
+ | <math>\therefore</math> <math>\angle PYX=\angle YXB</math> | ||
+ | |||
+ | <math>\therefore</math> <math>\sin \angle PYX=\sin \angle YXB=\sin \angle YXC=\sin \angle PYA</math> | ||
+ | |||
+ | Let the incircle of <math>ABC</math> be tangent to <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math>, let <math>MP=YP=x</math> and <math>NQ=YQ=y</math>. | ||
+ | |||
+ | Use Law of Sine in <math>\triangle APY</math> and <math>\triangle AXB</math>, we have | ||
+ | |||
+ | <math>\frac{\sin \angle PAY}{PY}=\frac{\sin \angle PYA}{PA}</math> | ||
+ | |||
+ | <math>\frac{\sin \angle BAX}{BX}=\frac{\sin \angle AXB}{AB}</math> | ||
+ | |||
+ | therefore we have | ||
+ | |||
+ | <math>\frac{3}{x}=\frac{7}{4-x}</math> | ||
+ | |||
+ | Solve this equation, we have <math>x=\frac{6}{5}</math> | ||
+ | |||
+ | As a result, <math>MB=4-x=\frac{14}{5}=BX</math>, <math>AM=x+3=\frac{21}{5}=AN</math>, <math>NC=8-AN=\frac{19}{5}=XC</math>, <math>AQ=\frac{21}{5}-y</math>, <math>PQ=\frac{6}{5}+y</math> | ||
+ | |||
+ | So, <math>BC=\frac{14}{5}+\frac{19}{5}=\frac{33}{5}</math> | ||
+ | |||
+ | Use Law of Cosine in <math>\triangle BAC</math> and <math>\triangle PAQ</math>, we have | ||
+ | |||
+ | <math>\cos \angle BAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}</math> | ||
+ | |||
+ | <math>\cos \angle PAQ=\frac{AP^2+AQ^2-PQ^2}{2\cdot AP\cdot AQ}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}-y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}</math> | ||
+ | |||
+ | And we have | ||
+ | |||
+ | <math>\cos \angle BAC=\cos \angle PAQ</math> | ||
+ | |||
+ | So | ||
+ | |||
+ | <math>\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}-y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}</math> | ||
+ | |||
+ | Solve this equation, we have <math>y=\frac{399}{295}=QN</math> | ||
+ | |||
+ | As a result, <math>AQ=AN-QN=\frac{21}{5}-\frac{399}{295}=\frac{168}{59}</math> | ||
+ | |||
+ | So, the final answer of this question is <math>168+59=\boxed {227}</math> | ||
+ | |||
{{AIME box|year=2018|n=II|num-b=13|num-a=15}} | {{AIME box|year=2018|n=II|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 05:29, 22 August 2018
Contents
Problem
The incircle of triangle is tangent to at . Let be the other intersection of with . Points and lie on and , respectively, so that is tangent to at . Assume that , , , and , where and are relatively prime positive integers. Find .
Solution 1
Let sides and be tangent to at and , respectively. Let and . Because and are both tangent to and and subtend the same arc of , it follows that . By equal tangents, . Applying the Law of Sines to yields Similarly, applying the Law of Sines to gives It follows that implying . Applying the same argument to yields from which . The requested sum is .
Solution 2 (Projective)
Let the incircle of be tangent to and at and . By Brianchon's theorem on tangential hexagons and , we know that and are concurrent at a point . Let . Then by La Hire's lies on the polar of so lies on the polar of . Therefore, also passes through . Then projecting through , we have Therefore, . Since we know that and . Therefore, and . Since , we also have . Solving for , we obtain . -Vfire
Solution 3 (Combination of Law of Sine and Law of Cosine)
Let the center of the incircle of be . Link and . Then we have
Let the incircle of be tangent to and at and , let and .
Use Law of Sine in and , we have
therefore we have
Solve this equation, we have
As a result, , , , ,
So,
Use Law of Cosine in and , we have
And we have
So
Solve this equation, we have
As a result,
So, the final answer of this question is
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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