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 "2016 AIME I Problems/Problem 6"

(→‎See also)
Line 2: Line 2:
 
In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>AB</math> at <math>L</math>. The line through <math>C</math> and <math>L</math> intersects the circumscribed circle of <math>\triangle ABC</math> at the two points <math>C</math> and <math>D</math>. If <math>LI=2</math> and <math>LD=3</math>, then <math>IC=\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
 
In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>AB</math> at <math>L</math>. The line through <math>C</math> and <math>L</math> intersects the circumscribed circle of <math>\triangle ABC</math> at the two points <math>C</math> and <math>D</math>. If <math>LI=2</math> and <math>LD=3</math>, then <math>IC=\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
  
−
==Solution==
+
=Solution=
 +
==Solution 1==
 
It is well known that <math>DA = DI = DB</math> and so we have <math>DA = DB = 5</math>. Then <math>\triangle DLB \sim \triangle ALC</math> and so <math>\frac{AL}{AC} = \frac{3}{5}</math> and from the angle bisector theorem <math>\frac{CI}{IL} = \frac{5}{3}</math> so <math>CI = \frac{10}{3}</math> and our answer is <math>\boxed{013}</math>
 
It is well known that <math>DA = DI = DB</math> and so we have <math>DA = DB = 5</math>. Then <math>\triangle DLB \sim \triangle ALC</math> and so <math>\frac{AL}{AC} = \frac{3}{5}</math> and from the angle bisector theorem <math>\frac{CI}{IL} = \frac{5}{3}</math> so <math>CI = \frac{10}{3}</math> and our answer is <math>\boxed{013}</math>
 +
 +
==Solution 2==
 +
 +
This is a cheap solution.
 +
 +
WLOG assume <math>\triangle ABC</math> is isosceles. Then, <math>L</math> is the midpoint of <math>AB</math>, and <math>\angle CLB=\angle CLA=90^\circ</math>. Draw the perpendicular from <math>I</math> to <math>CB</math>, and let it meet <math>CB</math> at <math>E</math>. Since <math>IL=2</math>, <math>IE</math> is also <math>2</math> (they are both inradii). Set <math>BD</math> as <math>x</math>. Then, triangles <math>BLD</math> and <math>CEI</math> are similar, and <math>\tfrac{2}{3}=\tfrac{CI}{x}</math>. Thus, <math>CI=\tfrac{2x}{3}</math>. <math>\triangle CBD~\triangle CEI</math>, so <math>\tfrac{IE}{DB}=\tfrac{CI}{CD}</math>. Thus <math>\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}</math>. Solving for <math>x</math>, we have:
 +
<math>x^2-2x-15=0</math>, or <math>x=5, -3</math>. <math>x</math> is positive, so <math>x=5</math>. As a result, <math>CI=\tfrac{2x}{3}=\tfrac{10}{3}</math> and the answer is <math>\boxed{013}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2016|n=I|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:20, 4 March 2016

Problem

In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$. The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI=2$ and $LD=3$, then $IC=\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

Solution 1

It is well known that $DA = DI = DB$ and so we have $DA = DB = 5$. Then $\triangle DLB \sim \triangle ALC$ and so $\frac{AL}{AC} = \frac{3}{5}$ and from the angle bisector theorem $\frac{CI}{IL} = \frac{5}{3}$ so $CI = \frac{10}{3}$ and our answer is $\boxed{013}$

Solution 2

This is a cheap solution.

WLOG assume $\triangle ABC$ is isosceles. Then, $L$ is the midpoint of $AB$, and $\angle CLB=\angle CLA=90^\circ$. Draw the perpendicular from $I$ to $CB$, and let it meet $CB$ at $E$. Since $IL=2$, $IE$ is also $2$ (they are both inradii). Set $BD$ as $x$. Then, triangles $BLD$ and $CEI$ are similar, and $\tfrac{2}{3}=\tfrac{CI}{x}$. Thus, $CI=\tfrac{2x}{3}$. $\triangle CBD~\triangle CEI$, so $\tfrac{IE}{DB}=\tfrac{CI}{CD}$. Thus $\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}$. Solving for $x$, we have: $x^2-2x-15=0$, or $x=5, -3$. $x$ is positive, so $x=5$. As a result, $CI=\tfrac{2x}{3}=\tfrac{10}{3}$ and the answer is $\boxed{013}$

See also

2016 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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=2016_AIME_I_Problems/Problem_6&oldid=77265"