Revision as of 00:19, 29 June 2020

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

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)

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

@@ Line 60: / Line 60: @@
 ==Solution 7==
-We can just say that quadrilateral <math>ADBC</math> is a right kite with right angles at <math>A</math> and <math>B</math>. Let us construct another similar right kite with the points of tangency on <math>AC</math> and <math>BC</math> called <math>E</math> and <math>F</math> respectively, point <math>I</math>, and point <math>C</math>. Note that we only have to look at one half of the circle since the diagram is not symmetrical. Let us call <math>CI</math> <math>x</math> for simplicity's sake. Based on the fact that <math>\triangle BCD</math> is similar to <math>\triangle FCI</math> we can use triangle proportionality to say that <math>BD</math> is <math>2\frac{x+5}{x}</math>. Using geometric mean theorem we can show that <math>BL</math> must be <math>\sqrt{3x+6}</math>. With Pythagorean Theorem we can say that <math>3x+6+9=4{(\frac{x+5}{x})}^2</math>. Multiplying both sides by <math>x^2</math> and moving everything to LHS will give you <math>3{x}^3+11{x}^2-40x-100=0</math> Since <math>x</math> must be in the form <math>\frac{p}{q}</math> we can assume that <math>x</math> is most likely a positive fraction in the form <math>\frac{p}{3}</math> where <math>p</math> is a factor of <math>100</math>. Testing the factors in synthetic division would lead <math>x = \frac{10}{3}</math>, giving us our desired answer <math>\boxed{013}</math>. ~Lopkiloinm
+We can just say that quadrilateral <math>ADBC</math> is a right kite with right angles at <math>A</math> and <math>B</math>. Let us construct another similar right kite with the points of tangency on <math>AC</math> and <math>BC</math> called <math>E</math> and <math>F</math> respectively, point <math>I</math>, and point <math>C</math>. Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call <math>CI</math> <math>x</math> for simplicity's sake. Based on the fact that <math>\triangle BCD</math> is similar to <math>\triangle FCI</math> we can use triangle proportionality to say that <math>BD</math> is <math>2\frac{x+5}{x}</math>. Using geometric mean theorem we can show that <math>BL</math> must be <math>\sqrt{3x+6}</math>. With Pythagorean Theorem we can say that <math>3x+6+9=4{(\frac{x+5}{x})}^2</math>. Multiplying both sides by <math>x^2</math> and moving everything to LHS will give you <math>3{x}^3+11{x}^2-40x-100=0</math> Since <math>x</math> must be in the form <math>\frac{p}{q}</math> we can assume that <math>x</math> is most likely a positive fraction in the form <math>\frac{p}{3}</math> where <math>p</math> is a factor of <math>100</math>. Testing the factors in synthetic division would lead <math>x = \frac{10}{3}</math>, giving us our desired answer <math>\boxed{013}</math>. ~Lopkiloinm
 == See also ==
 {{AIME box|year=2016|n=I|num-b=5|num-a=7}}
 {{MAA Notice}}

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