Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AIME II Problems/Problem 14"

(Created page with "==Problem== The incircle <math>\omega</math> of triangle <math>ABC</math> is tangent to <math>\overline{BC}</math> at <math>X</math>. Let <math>Y \neq X</math> be the other i...")
 
Line 3: Line 3:
 
The incircle <math>\omega</math> of triangle <math>ABC</math> is tangent to <math>\overline{BC}</math> at <math>X</math>. Let <math>Y \neq X</math> be the other intersection of <math>\overline{AX}</math> with <math>\omega</math>. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>\overline{PQ}</math> is tangent to <math>\omega</math> at <math>Y</math>. Assume that <math>AP = 3</math>, <math>PB = 4</math>, <math>AC = 8</math>, and <math>AQ = \dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
The incircle <math>\omega</math> of triangle <math>ABC</math> is tangent to <math>\overline{BC}</math> at <math>X</math>. Let <math>Y \neq X</math> be the other intersection of <math>\overline{AX}</math> with <math>\omega</math>. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>\overline{PQ}</math> is tangent to <math>\omega</math> at <math>Y</math>. Assume that <math>AP = 3</math>, <math>PB = 4</math>, <math>AC = 8</math>, and <math>AQ = \dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 +
 +
==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>.
 +
 +
==Solution 2 (Projective)==
 +
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>.
 +
-Vfire
 
{{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 08:16, 26 March 2018

Problem

The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$. Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$. Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$, respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$. Assume that $AP = 3$, $PB = 4$, $AC = 8$, and $AQ = \dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.


Solution 1

Let sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $Z$ and $W$, respectively. Let $\alpha = \angle BAX$ and $\beta = \angle AXC$. Because $\overline{PQ}$ and $\overline{BC}$ are both tangent to $\omega$ and $\angle YXC$ and $\angle QYX$ subtend the same arc of $\omega$, it follows that $\angle AYP = \angle QYX = \angle YXC = \beta$. By equal tangents, $PZ = PY$. Applying the Law of Sines to $\triangle APY$ yields \[\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.\]Similarly, applying the Law of Sines to $\triangle ABX$ gives \[\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.\]It follows that \[2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,\]implying $AZ = \tfrac{21}5$. Applying the same argument to $\triangle AQY$ yields \[2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),\]from which $AQ = \tfrac{168}{59}$. The requested sum is $168 + 59 = \boxed{227}$.

Solution 2 (Projective)

Let sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $Z$ and $W$, respectively. Let $\alpha = \angle BAX$ and $\beta = \angle AXC$. Because $\overline{PQ}$ and $\overline{BC}$ are both tangent to $\omega$ and $\angle YXC$ and $\angle QYX$ subtend the same arc of $\omega$, it follows that $\angle AYP = \angle QYX = \angle YXC = \beta$. By equal tangents, $PZ = PY$. Applying the Law of Sines to $\triangle APY$ yields \[\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.\]Similarly, applying the Law of Sines to $\triangle ABX$ gives \[\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.\]It follows that \[2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,\]implying $AZ = \tfrac{21}5$. Applying the same argument to $\triangle AQY$ yields \[2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),\]from which $AQ = \tfrac{168}{59}$. The requested sum is $168 + 59 = \boxed{227}$. -Vfire

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_14&oldid=93508"