Difference between revisions of "2018 AIME II Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
− | Let sides <math>\overline{AB}</math> and <math>\overline{AC}</math> be tangent to <math>\omega</math> at <math>Z</math> and <math>W</math>, respectively. Let <math>\alpha = \angle BAX</math> and <math>\beta = \angle AXC</math>. Because <math>\overline{PQ}</math> and <math>\overline{BC}</math> are both tangent to <math>\omega</math> and <math>\angle YXC</math> and <math>\angle QYX</math> subtend the same arc of <math>\omega</math>, it follows that <math>\angle AYP = \angle QYX = \angle YXC = \beta</math>. By equal tangents, <math>PZ = PY</math>. Applying the Law of Sines to <math>\triangle APY</math> yields <cmath>\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.</cmath>Similarly, applying the Law of Sines to <math>\triangle ABX</math> gives <cmath>\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.</cmath>It follows that <cmath>2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,</cmath>implying <math>AZ = \tfrac{21}5</math>. Applying the same argument to <math>\triangle AQY</math> yields <cmath>2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),</cmath>from which <math>AQ = \tfrac{168}{59}</math>. The requested sum is <math>168 + 59 = \boxed{227}</math>. | + | Let the sides <math>\overline{AB}</math> and <math>\overline{AC}</math> be tangent to <math>\omega</math> at <math>Z</math> and <math>W</math>, respectively. Let <math>\alpha = \angle BAX</math> and <math>\beta = \angle AXC</math>. Because <math>\overline{PQ}</math> and <math>\overline{BC}</math> are both tangent to <math>\omega</math> and <math>\angle YXC</math> and <math>\angle QYX</math> subtend the same arc of <math>\omega</math>, it follows that <math>\angle AYP = \angle QYX = \angle YXC = \beta</math>. By equal tangents, <math>PZ = PY</math>. Applying the Law of Sines to <math>\triangle APY</math> yields <cmath>\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.</cmath>Similarly, applying the Law of Sines to <math>\triangle ABX</math> gives <cmath>\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.</cmath>It follows that <cmath>2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,</cmath>implying <math>AZ = \tfrac{21}5</math>. Applying the same argument to <math>\triangle AQY</math> yields <cmath>2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),</cmath>from which <math>AQ = \tfrac{168}{59}</math>. The requested sum is <math>168 + 59 = \boxed{227}</math>. |
==Solution 2 (Projective)== | ==Solution 2 (Projective)== | ||
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>. By Brianchon's theorem on tangential hexagons <math>QNCBMP</math> and <math>PYQCXB</math>, we know that <math>MN,CP,BQ</math> and <math>XY</math> are concurrent at a point <math>O</math>. Let <math>PQ \cap BC = Z</math>. Then by La Hire's <math>A</math> lies on the polar of <math>Z</math> so <math>Z</math> lies on the polar of <math>A</math>. Therefore, <math>MN</math> also passes through <math>Z</math>. Then projecting through <math>Z</math>, we have | 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>. By Brianchon's theorem on tangential hexagons <math>QNCBMP</math> and <math>PYQCXB</math>, we know that <math>MN,CP,BQ</math> and <math>XY</math> are concurrent at a point <math>O</math>. Let <math>PQ \cap BC = Z</math>. Then by La Hire's <math>A</math> lies on the polar of <math>Z</math> so <math>Z</math> lies on the polar of <math>A</math>. Therefore, <math>MN</math> also passes through <math>Z</math>. Then projecting through <math>Z</math>, we have | ||
− | <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> | + | <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>MP = \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 | ||
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<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 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} | + | <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 | And we have | ||
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So | 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} | + | <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> | Solve this equation, we have <math>y=\frac{399}{295}=QN</math> | ||
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So, the final answer of this question is <math>168+59=\boxed {227}</math> | So, the final answer of this question is <math>168+59=\boxed {227}</math> | ||
+ | ~Solution by <math>BladeRunnerAUG</math> (Frank FYC) | ||
+ | |||
+ | ==See Also== | ||
{{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}} |
Latest revision as of 14:32, 12 March 2021
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 the 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
~Solution by (Frank FYC)
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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