Difference between revisions of "2016 AIME I Problems/Problem 6"

(Solution 8 (Cyclic Quadrilaterals))
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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{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</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{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
=Solution=
 
 
==Solution 1==
 
==Solution 1==
 
Suppose we label the angles as shown below.
 
Suppose we label the angles as shown below.
Line 47: Line 46:
 
==Solution 4==
 
==Solution 4==
  
Since <math>\angle{LAD} = \angle{BDC}</math> and <math>\angle{DLA}=\angle{DBC}</math>, <math>\triangle{DLA}\sim\triangle{DBC}</math>. Also, <math>\angle{DAC}=\angle{BLC}</math> and <math>\angle{ACD}=\angle{LCB}</math> so <math>\triangle{DAC}\sim\triangle{BLC}</math>. Now we can call <math>AC</math>, <math>b</math> and <math>BC</math>, <math>a</math>. By angle bisector theorem, <math>\frac{AD}{DB}=\frac{AC}{BC}</math>. So let <math>AD=bk</math> and <math>DB=ak</math> for some value of <math>k</math>. Now call <math>IC=x</math>. By the similar triangles we found earlier, <math>\frac{3}{ak}=\frac{bk}{x+2}</math> and <math>\frac{b}{x+5}=\frac{x+2}{a}</math>. We can simplify this to <math>abk^2=3x+6</math> and <math>ab=(x+5)(x+2)</math>. So we can plug the <math>ab</math> into the first equation and get <math>(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3</math>. We can now draw a line through <math>A</math> and <math>I</math> that intersects <math>BC</math> at <math>E</math>. By mass points, we can assign a mass of <math>a</math> to <math>A</math>, <math>b</math> to <math>B</math>, and <math>a+b</math> to <math>D</math>. We can also assign a mass of <math>(a+b)k</math> to <math>C</math> by angle bisector theorem. So the ratio of <math>\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}</math>. So since <math>k=\frac{2}{x}</math>, we can plug this back into the original equation to get <math>\left(\frac{2}{x}\right)^2(x+5)=3</math>. This means that <math>\frac{3x^2}{4}-x-5=0</math> which has roots -2 and <math>\frac{10}{3}</math> which means our <math>CI=\frac{10}{3}</math> and our answer is <math>\boxed{013}</math>.
+
Since <math>\angle{LAD} = \angle{BDC}</math> and <math>\angle{DLA}=\angle{DCB}</math>, <math>\triangle{DLA}\sim\triangle{DBC}</math>. Also, <math>\angle{DAC}=\angle{BLC}</math> and <math>\angle{ACD}=\angle{LCB}</math> so <math>\triangle{DAC}\sim\triangle{BLC}</math>. Now we can call <math>AC</math>, <math>b</math> and <math>BC</math>, <math>a</math>. By angle bisector theorem, <math>\frac{AD}{DB}=\frac{AC}{BC}</math>. So let <math>AD=bk</math> and <math>DB=ak</math> for some value of <math>k</math>. Now call <math>IC=x</math>. By the similar triangles we found earlier, <math>\frac{3}{ak}=\frac{bk}{x+2}</math> and <math>\frac{b}{x+5}=\frac{x+2}{a}</math>. We can simplify this to <math>abk^2=3x+6</math> and <math>ab=(x+5)(x+2)</math>. So we can plug the <math>ab</math> into the first equation and get <math>(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3</math>. We can now draw a line through <math>A</math> and <math>I</math> that intersects <math>BC</math> at <math>E</math>. By mass points, we can assign a mass of <math>a</math> to <math>A</math>, <math>b</math> to <math>B</math>, and <math>a+b</math> to <math>D</math>. We can also assign a mass of <math>(a+b)k</math> to <math>C</math> by angle bisector theorem. So the ratio of <math>\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}</math>. So since <math>k=\frac{2}{x}</math>, we can plug this back into the original equation to get <math>\left(\frac{2}{x}\right)^2(x+5)=3</math>. This means that <math>\frac{3x^2}{4}-x-5=0</math> which has roots -2 and <math>\frac{10}{3}</math> which means our <math>CI=\frac{10}{3}</math> and our answer is <math>\boxed{013}</math>.
  
 
==Solution 5==
 
==Solution 5==
Line 56: Line 55:
 
~Pluto1708
 
~Pluto1708
  
Alternate solution: "We can use the angle bisector theorem on <math>\triangle CBL</math> and bisector <math>BI</math> to get that <math>\tfrac{CI}{IL}=\tfrac{CI}{2}=\tfrac{BC}{BL}</math>. Since <math>\triangle CBL \sim \triangle ADL</math>, we get <math>\tfrac{BC}{BL}=\tfrac{AD}{DL}=\tfrac{5}{3}</math>. Thus, <math>CI=\tfrac{10}{3}</math> and <math>p+q=\boxed{13}</math>."
+
Alternate solution: We can use the angle bisector theorem on <math>\triangle CBL</math> and bisector <math>BI</math> to get that <math>\tfrac{CI}{IL}=\tfrac{CI}{2}=\tfrac{BC}{BL}</math>. Since <math>\triangle CBL \sim \triangle ADL</math>, we get <math>\tfrac{BC}{BL}=\tfrac{AD}{DL}=\tfrac{5}{3}</math>. Thus, <math>CI=\tfrac{10}{3}</math> and <math>p+q=\boxed{13}</math>.
 
(https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)
 
(https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)
  
Line 64: Line 63:
 
==Solution 8 (Cyclic Quadrilaterals)==
 
==Solution 8 (Cyclic Quadrilaterals)==
  
First, using the diagram in solution one as a reference, extend the angle bisector of angle <math>\angle ACB</math>
+
<asy>
 +
size(150);
 +
import olympiad;
 +
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
 +
pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2));
 +
pair C=intersectionpoints(circle(A,b),circle(B,a))[0];
 +
pair I=incenter(A,B,C);
 +
pair L=extension(C,D,A,B);
 +
dot(I^^A^^B^^C^^D);
 +
draw(C--D);
 +
path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
 +
draw(A--B--D--cycle);
 +
draw(circumcircle(A,B,D));
 +
draw(A--C--B);
 +
draw(A--I--B^^C--I);
 +
draw(incircle(A,B,C));
 +
label("$A$",A,SW,fontsize(8));
 +
label("$B$",B,SE,fontsize(8));
 +
label("$C$",C,N,fontsize(8));
 +
label("$D$",D,S,fontsize(8));
 +
label("$I$",I,NE,fontsize(8));
 +
label("$L$",L,SW,fontsize(8));
 +
label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8));
 +
label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8));
 +
label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8));
 +
label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8));
 +
label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8));
 +
label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8));
 +
</asy>
 +
Connect <math>D</math> to <math>A</math> and <math>D</math> to <math>B</math> to form quadrilateral <math>ACBD</math>. Since quadrilateral <math>ACBD</math> is cyclic, we can apply Ptolemy's Theorem on the quadrilateral.
 +
 
 +
Denote the length of <math>BD</math> and <math>AD</math> as <math>z</math> (they must be congruent, as <math>\angle ABD</math> and <math>\angle DAB</math> are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at <math>D</math>), and the lengths of <math>BC</math>, <math>AC</math>, <math>AB</math>, and <math>CI</math> as <math>a,b,c, x</math>, respectively.
 +
 
 +
After applying Ptolemy's, one will get that:
 +
 
 +
<cmath>z(a+b)=c(x+5)</cmath>
 +
 
 +
Next, since <math>ACBD</math> is cyclic, triangles <math>ALD</math> and <math>CLB</math> are similar, yielding the following equation once simplifications are made to the equation <math>\frac{AD}{CB}=\frac{AL}{BL}</math>, with the length of <math>BL</math> written in terms of <math>a,b,c</math> using the angle bisector theorem on triangle <math>ABC</math>:
 +
 
 +
<cmath>zc=3(a+b)</cmath>
 +
 
 +
Next, drawing in the bisector of <math>\angle BAC</math> to the incenter <math>I</math>, and applying the angle bisector theorem, we have that:
 +
 
 +
<cmath>cx=2(a+b)</cmath>
 +
 
 +
Now, solving for <math>z</math> in the second equation, and <math>x</math> in the third equation and plugging them both back into the first equation, and making the substitution <math>w=\frac{a+b}{c}</math>, we get the quadratic equation:
 +
 
 +
<cmath>3w^2-2w-5=0</cmath>
 +
 
 +
Solving, we get <math>w=5/3</math>, which gives <math>z=5</math> and <math>x=10/3</math>, when we rewrite the above equations in terms of <math>w</math>. Thus, our answer is <math>\boxed{013}</math> and we're done.
 +
 
 +
-mathislife52
 +
 
 +
==Solution 9(Visual)==
 +
[[File:2016 AIME I 6b.png|500px]]
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Solution 10==
 +
Let <math>AB=c,BC=a,CA=b</math>, and <math>x=\tfrac{a+b}{c}</math>. Then, notice that <math>\tfrac{CI}{IL}=\tfrac{a+b}{c}=x</math>, so <math>CI=IL\cdot{}x=2x</math>. Also, by the incenter-excenter lemma, <math>AD=BD=ID=IL+LD=5</math>. Therefore, by Ptolemy's Theorem on cyclic quadrilateral <math>ABCD</math>, <math>5a+5b=c(2x+5)</math>, so <math>5\left(\tfrac{a+b}{c}\right)=2x+5</math>, so <math>5x=2x+5</math>. Solving, we get that <math>x=\tfrac{5}{3}</math>, so <math>CI=\tfrac{10}{3}</math> and the answer is <math>10+3=\boxed{013}</math>.
 +
 
 +
==Solution 11==
 +
 
 +
Perform a <math>\sqrt{bc}</math> Inversion followed by a reflection along the angle bisector of <math>\angle BCA</math>.
 +
 
 +
It's well known that
 +
<cmath>AB \leftrightarrow \odot CBA \implies L \leftrightarrow D</cmath>
 +
<cmath>I \leftrightarrow I_A</cmath>
 +
where <math>I_A</math> is the <math>A-</math>excenter.
 +
 
 +
Also by Fact 5, <math>DI_A = 5</math>.
 +
 
 +
So,
 +
<cmath>CL \cdot CD = CI \cdot CI_A</cmath>
 +
<cmath>\implies (CI + IL) \cdot (CI + ID) = (CI) \cdot (CI + II_A)</cmath>
 +
<cmath>\implies (CI + 2) \cdot (CI + 5) = (CI) \cdot (CI + 10)</cmath>
 +
<cmath>\implies 7CI +10= 10CI</cmath>
 +
<cmath>\implies CI = \boxed{\dfrac{10}{3}}.\blacksquare</cmath>
 +
 
 +
~kamatadu
 +
 
 +
 
 +
==Solution 12==
 +
Without loss of generality, let <math>\triangle ABC</math> be isosceles. Note that by the incenter-excenter lemma, <math>DI = DA = DB.</math> Hence, <math>DA=DB=5.</math> Let the point of tangency of the incircle and <math>\overline{BC}</math> be <math>F</math> and the point of tangency of the incircle and <math>\overline{AC}</math> be <math>E.</math> We note that <math>\angle ALC = \angle BLC = 90^\circ</math> and <math>LA=LB=4,</math> which immediately gives <math>AE=BF=4.</math> Applying the Pythagorean Theorem on <math>\triangle ALC</math> and <math>\triangle IEC</math> gives <math>2^2+x^2=y^2</math> and <math>4^2+(2+y)^2 = (4+x)^2.</math> Solving for <math>y</math> gives us <math>y=\frac{10}{3}.</math> Therefore, <math>IC = \frac{10}{3}</math> so the answer is <math>\boxed{13}.</math>  
 +
 
 +
~peelybonehead
  
 
== 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}}

Latest revision as of 20:04, 27 August 2023

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 $IC=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

Suppose we label the angles as shown below. [asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label("$A$",A,SW,fontsize(8)); label("$B$",B,SE,fontsize(8)); label("$C$",C,N,fontsize(8)); label("$D$",D,S,fontsize(8)); label("$I$",I,NE,fontsize(8)); label("$L$",L,SW,fontsize(8)); label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy] As $\angle BCD$ and $\angle BAD$ intercept the same arc, we know that $\angle BAD=\gamma$. Similarly, $\angle ABD=\gamma$. Also, using $\triangle ICA$, we find $\angle CIA=180-\alpha-\gamma$. Therefore, $\angle AID=\alpha+\gamma$. Therefore, $\angle DAI=\angle AID=\alpha+\gamma$, so $\triangle AID$ must be isosceles with $AD=ID=5$. Similarly, $BD=ID=5$. Then $\triangle DLB \sim \triangle ALC$, hence $\frac{AL}{AC} = \frac{3}{5}$. Also, $AI$ bisects $\angle LAC$, so by the Angle Bisector Theorem $\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}$. Thus $CI = \frac{10}{3}$, and the answer is $\boxed{013}$.

Solution 2

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, $CI=\tfrac{2x}{3}$. $\triangle CBD \sim \triangle CEI$, 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 circumradius, and $r$ the inradius. A simple sketch will reveal that $\triangle ABC$ must 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 $R+d=DL+LI=5$. 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 $\boxed{013}$.

Solution 4

Since $\angle{LAD} = \angle{BDC}$ and $\angle{DLA}=\angle{DCB}$, $\triangle{DLA}\sim\triangle{DBC}$. Also, $\angle{DAC}=\angle{BLC}$ and $\angle{ACD}=\angle{LCB}$ so $\triangle{DAC}\sim\triangle{BLC}$. Now we can call $AC$, $b$ and $BC$, $a$. By angle bisector theorem, $\frac{AD}{DB}=\frac{AC}{BC}$. So let $AD=bk$ and $DB=ak$ for some value of $k$. Now call $IC=x$. By the similar triangles we found earlier, $\frac{3}{ak}=\frac{bk}{x+2}$ and $\frac{b}{x+5}=\frac{x+2}{a}$. We can simplify this to $abk^2=3x+6$ and $ab=(x+5)(x+2)$. So we can plug the $ab$ into the first equation and get $(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3$. We can now draw a line through $A$ and $I$ that intersects $BC$ at $E$. By mass points, we can assign a mass of $a$ to $A$, $b$ to $B$, and $a+b$ to $D$. We can also assign a mass of $(a+b)k$ to $C$ by angle bisector theorem. So the ratio of $\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}$. So since $k=\frac{2}{x}$, we can plug this back into the original equation to get $\left(\frac{2}{x}\right)^2(x+5)=3$. This means that $\frac{3x^2}{4}-x-5=0$ which has roots -2 and $\frac{10}{3}$ which means our $CI=\frac{10}{3}$ and our answer is $\boxed{013}$.

Solution 5

Since $\angle BCD$ and $\angle BAD$ both intercept arc $BD$, it follows that $\angle BAD=\gamma$. Note that $\angle AID=\alpha+\gamma$ by the external angle theorem. It follows that $\angle DAI=\angle AID=\alpha+\gamma$, so we must have that $\triangle AID$ is isosceles, yielding $AD=ID=5$. Note that $\triangle DLA \sim \triangle DAC$, so $\frac{DA}{DL} = \frac{DC}{DA}$. This yields $DC = \frac{25}{3}$. It follows that $CI = DC - DI = \frac{10}{3}$, giving a final answer of $\boxed{013}$.

Solution 6

Let $I_C$ be the excenter opposite to $C$ in $ABC$. By the incenter-excenter lemma $DI=DC \therefore$ $LI_C=8,LI=2,II_C=10$. Its well known that $(I_C,I,L,C)=-1 \implies \dfrac{LI_C}{LI}.\dfrac{CI}{CI_C}=-1 \implies \dfrac{CI}{CI+10}=\dfrac{1}{4} \implies \boxed{CI=\dfrac{10}{3}}$.$\blacksquare$ ~Pluto1708

Alternate solution: We can use the angle bisector theorem on $\triangle CBL$ and bisector $BI$ to get that $\tfrac{CI}{IL}=\tfrac{CI}{2}=\tfrac{BC}{BL}$. Since $\triangle CBL \sim \triangle ADL$, we get $\tfrac{BC}{BL}=\tfrac{AD}{DL}=\tfrac{5}{3}$. Thus, $CI=\tfrac{10}{3}$ and $p+q=\boxed{13}$. (https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)

Solution 7

We can just say that quadrilateral $ADBC$ is a right kite with right angles at $A$ and $B$. Let us construct another similar right kite with the points of tangency on $AC$ and $BC$ called $E$ and $F$ respectively, point $I$, and point $C$. Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call $CI$ $x$ for simplicity's sake. Based on the fact that $\triangle BCD$ is similar to $\triangle FCI$ we can use triangle proportionality to say that $BD$ is $2\frac{x+5}{x}$. Using geometric mean theorem we can show that $BL$ must be $\sqrt{3x+6}$. With Pythagorean Theorem we can say that $3x+6+9=4{(\frac{x+5}{x})}^2$. Multiplying both sides by $x^2$ and moving everything to LHS will give you $3{x}^3+11{x}^2-40x-100=0$ Since $x$ must be in the form $\frac{p}{q}$ we can assume that $x$ is most likely a positive fraction in the form $\frac{p}{3}$ where $p$ is a factor of $100$. Testing the factors in synthetic division would lead $x = \frac{10}{3}$, giving us our desired answer $\boxed{013}$. ~Lopkiloinm

Solution 8 (Cyclic Quadrilaterals)

[asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label("$A$",A,SW,fontsize(8)); label("$B$",B,SE,fontsize(8)); label("$C$",C,N,fontsize(8)); label("$D$",D,S,fontsize(8)); label("$I$",I,NE,fontsize(8)); label("$L$",L,SW,fontsize(8)); label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy] Connect $D$ to $A$ and $D$ to $B$ to form quadrilateral $ACBD$. Since quadrilateral $ACBD$ is cyclic, we can apply Ptolemy's Theorem on the quadrilateral.

Denote the length of $BD$ and $AD$ as $z$ (they must be congruent, as $\angle ABD$ and $\angle DAB$ are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at $D$), and the lengths of $BC$, $AC$, $AB$, and $CI$ as $a,b,c, x$, respectively.

After applying Ptolemy's, one will get that:

\[z(a+b)=c(x+5)\]

Next, since $ACBD$ is cyclic, triangles $ALD$ and $CLB$ are similar, yielding the following equation once simplifications are made to the equation $\frac{AD}{CB}=\frac{AL}{BL}$, with the length of $BL$ written in terms of $a,b,c$ using the angle bisector theorem on triangle $ABC$:

\[zc=3(a+b)\]

Next, drawing in the bisector of $\angle BAC$ to the incenter $I$, and applying the angle bisector theorem, we have that:

\[cx=2(a+b)\]

Now, solving for $z$ in the second equation, and $x$ in the third equation and plugging them both back into the first equation, and making the substitution $w=\frac{a+b}{c}$, we get the quadratic equation:

\[3w^2-2w-5=0\]

Solving, we get $w=5/3$, which gives $z=5$ and $x=10/3$, when we rewrite the above equations in terms of $w$. Thus, our answer is $\boxed{013}$ and we're done.

-mathislife52

Solution 9(Visual)

2016 AIME I 6b.png vladimir.shelomovskii@gmail.com, vvsss

Solution 10

Let $AB=c,BC=a,CA=b$, and $x=\tfrac{a+b}{c}$. Then, notice that $\tfrac{CI}{IL}=\tfrac{a+b}{c}=x$, so $CI=IL\cdot{}x=2x$. Also, by the incenter-excenter lemma, $AD=BD=ID=IL+LD=5$. Therefore, by Ptolemy's Theorem on cyclic quadrilateral $ABCD$, $5a+5b=c(2x+5)$, so $5\left(\tfrac{a+b}{c}\right)=2x+5$, so $5x=2x+5$. Solving, we get that $x=\tfrac{5}{3}$, so $CI=\tfrac{10}{3}$ and the answer is $10+3=\boxed{013}$.

Solution 11

Perform a $\sqrt{bc}$ Inversion followed by a reflection along the angle bisector of $\angle BCA$.

It's well known that \[AB \leftrightarrow \odot CBA \implies L \leftrightarrow D\] \[I \leftrightarrow I_A\] where $I_A$ is the $A-$excenter.

Also by Fact 5, $DI_A = 5$.

So, \[CL \cdot CD = CI \cdot CI_A\] \[\implies (CI + IL) \cdot (CI + ID) = (CI) \cdot (CI + II_A)\] \[\implies (CI + 2) \cdot (CI + 5) = (CI) \cdot (CI + 10)\] \[\implies 7CI +10= 10CI\] \[\implies CI = \boxed{\dfrac{10}{3}}.\blacksquare\]

~kamatadu


Solution 12

Without loss of generality, let $\triangle ABC$ be isosceles. Note that by the incenter-excenter lemma, $DI = DA = DB.$ Hence, $DA=DB=5.$ Let the point of tangency of the incircle and $\overline{BC}$ be $F$ and the point of tangency of the incircle and $\overline{AC}$ be $E.$ We note that $\angle ALC = \angle BLC = 90^\circ$ and $LA=LB=4,$ which immediately gives $AE=BF=4.$ Applying the Pythagorean Theorem on $\triangle ALC$ and $\triangle IEC$ gives $2^2+x^2=y^2$ and $4^2+(2+y)^2 = (4+x)^2.$ Solving for $y$ gives us $y=\frac{10}{3}.$ Therefore, $IC = \frac{10}{3}$ so the answer is $\boxed{13}.$

~peelybonehead

See also

2016 AIME I (ProblemsAnswer KeyResources)
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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