2016 AIME I Problems/Problem 6

Revision as of 23:17, 25 September 2019 by Mag1c (talk | contribs) (Solution 6: alternate solution)


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 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 $\angle{DLA}=\angle{DBC}$, $\triangle{DLA}\sim\triangle{DBC}$. 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)

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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