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

(Solution 7)
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==Problem==
 
==Problem==
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>.
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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>.
  
 
=Solution=
 
=Solution=
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(https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)
 
(https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)
  
== Solution 7 ==
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==Solution 7==
 
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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
First, we know that <math>AD = ID = BD = 5</math> by the incenter-excenter lemma (Fact 5). If you are not familiar with this lemma, it can be pretty useful in some problems so it might be a good chance to get acquainted with it. :)
 
 
 
Now because we are dealing with circumcircles and angle bisectors, we try to solve the problem using similar triangles and the angle bisector theorem.
 
 
 
(Another cool fact is because <math>CD</math> bisects angle <math>ACB</math>, <math>D</math> is the midpoint of arc <math>AB</math>.)
 
 
 
Therefore, we know that <math><ABD = <ACD = \alpha</math>, and by similar reasoning, <math><BAD = <DCB = \alpha</math>. (Note: I let <math>\alpha = <ACD = <BCD</math>).
 
 
 
So we try now to exploit that we have 2 pairs of equal triangles:
 
 
 
(First, let <math>CI = x</math>)
 
Triangle <math>BLD</math> is similar to triangle <math>CLA</math>: <math>\frac{BL}{5} = \frac{x + 2}{AC}</math>
 
Triangle <math>ALD</math> is similar to triangle <math>CLB</math>: <math>\frac{AL}{5} = \frac{x + 2}{BC}</math>
 
 
 
But they don’t really help us. Hm....
 
 
 
Well, usually when we have an in center of a triangle, it is usually good to connect all the vertices pod the triangle to the incenter, so let’s try that. Maybe then we can apply the angle bisected theorem then!
 
 
 
First connect <math>A</math> to <math>I</math> and <math>B</math> to <math>I</math>.
 
Then by the angle bisector theorem, <math>AL/AC = \frac{2}{x}</math>.
 
Wait! Since triangle <math>ALC</math> and <math>DLB</math> are similar, we have <math>\frac{AL}{AC} = \frac{DL}{DB}</math>.
 
And therefore, <math>\frac{2}{x} = \frac{3}{5}</math>, so <math>3x = 10</math>, therefore, <math>x = 10/3</math>, and so <math>m + n = 10 + 3 = 13.</math>
 
 
 
Key Concepts:
 
#6 usually isn’t too hard a problem on the AIME’s. The method usually is around 3 lines. Maybe not all the work, but the method shouldn’t be too bad.
 
 
 
When dealing with intersecting circles, or circles in general, always remember similar triangles! Try angle chasing!
 
 
 
On top of that, when dealing with angle bisectors, think about using the angle bisector theorem and similar triangles. Circles + angles bisectors usually means similar triangles!
 
 
 
(Professor-Mom)
 
  
 
== 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 16:55, 25 August 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 $IC=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

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

See also

2016 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions

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