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Difference between revisions of "2016 AIME I Problems/Problem 6"
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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:
 
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>
 
<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>
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 +
==Solution 3==
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 +
WLOG assume <math>\triangle ABC</math> is isosceles (with vertex <math>C</math>). Let <math>O</math> be the center of the circumcircle, <math>r</math> the inradius, and <math>R</math> the circumradius. A simple sketch will reveal that <math>\triangle ABC</math> must also be obtuse (as an acute triangle will result in <math>LI</math> being greater than <math>DL</math>) and that <math>O</math> and <math>I</math> are collinear. Next, if <math>OI=d</math>, <math>DO+OI=R+d</math> and <math>R+d=DL+DI=5</math>. Euler gives us that <math>d^{2}=R(R-2r)</math>, and in this case, <math>r=LI=2</math>. Thus, <math>d=\sqrt{R^{2}-4R}</math>. Solving for <math>d</math>, we have <math>R+\sqrt{R^{2}-4R}=5</math>, then <math>R^{2}-4R=25-10R+R^{2}</math>, yielding <math>R=\frac{25}{6}</math>. Next, <math>R+d=5</math> so <math>d=\frac{5}{6}</math>. Finally, <math>OC=OI+IC</math> gives us <math>R=d+IC</math>, and <math>IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}</math>. Our answer is then <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 19:39, 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 $\boxed{013}$

Solution 3

WLOG assume $\triangle ABC$ is isosceles (with vertex $C$). Let $O$ be the center of the circumcircle, $r$ the inradius, and $R$ the circumradius. A simple sketch will reveal that $\triangle ABC$ must also be obtuse (as an acute triangle will result in $LI$ being greater than $DL$) and that $O$ and $I$ are collinear. Next, if $OI=d$, $DO+OI=R+d$ and $R+d=DL+DI=5$. Euler gives us that $d^{2}=R(R-2r)$, and in this case, $r=LI=2$. Thus, $d=\sqrt{R^{2}-4R}$. Solving for $d$, we have $R+\sqrt{R^{2}-4R}=5$, then $R^{2}-4R=25-10R+R^{2}$, yielding $R=\frac{25}{6}$. Next, $R+d=5$ so $d=\frac{5}{6}$. Finally, $OC=OI+IC$ gives us $R=d+IC$, and $IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}$. Our answer is then $\boxed{013}$.

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

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