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Difference between revisions of "2011 AIME I Problems/Problem 4"

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== Solution 2 ==
 
== Solution 2 ==
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Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>IM \perp MC</math> and <math>IN \perp NC</math>, we have <math>CMIN</math> is a cyclic quadrilateral. Consequently, <math>\frac{MN}{\sin \angle MIN} = 2R = CI</math>. Since <math>\sin \angle MIN = \sin (90 - \frac{\angle BAC}{2}) = \cos \angle IAK</math>, we have that <math>MN = AI \cdot \cos \angle IAK</math>. Letting <math>X</math> be the point of contact of the incircle of <math>ABC</math> with side <math>AC</math>, we have <math>AX=MN</math>. Thus, <math>MN=\frac{117+120-125}{2}=\boxed{056}</math>
+
Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>IM \perp MC</math> and <math>IN \perp NC</math>, we have <math>CMIN</math> is a cyclic quadrilateral. Consequently, <math>\frac{MN}{\sin \angle MIN} = 2R = CI</math>. Since <math>\sin \angle MIN = \sin (90^\circ - \tfrac 12 \angle BAC) = \cos \angle IAK</math>, we have that <math>MN = AI \cdot \cos \angle IAK</math>. Letting <math>X</math> be the point of contact of the incircle of <math>ABC</math> with side <math>AC</math>, we have <math>AX=MN</math>. Thus, <math>MN=\frac{117+120-125}{2}=\boxed{056}</math>
  
 
== Solution 3 (Bash) ==  
 
== Solution 3 (Bash) ==  

Revision as of 12:08, 3 August 2022

Problem

In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find $MN$.

Solution 1

Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$, respectively. [asy] defaultpen(fontsize(10)+0.8); size(200); pen p=fontsize(9)+linewidth(3); pair A,B,C,D,K,L,M,N,P,Q; A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L); draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5); dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p); label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10)); [/asy] Since ${BM}$ is the angle bisector of angle $B$, and ${CM}$ is perpendicular to ${BM}$, so $BP=BC=120$, and $M$ is the midpoint of ${CP}$. For the same reason, $AQ=AC=117$, and $N$ is the midpoint of ${CQ}$. Hence $MN=\tfrac 12 PQ$. Since \[PQ=BP+AQ-AB=120+117-125=112,\] so $MN=\boxed{056}$.

Solution 2

Let $I$ be the incenter of $ABC$. Now, since $IM \perp MC$ and $IN \perp NC$, we have $CMIN$ is a cyclic quadrilateral. Consequently, $\frac{MN}{\sin \angle MIN} = 2R = CI$. Since $\sin \angle MIN = \sin (90^\circ - \tfrac 12 \angle BAC) = \cos \angle IAK$, we have that $MN = AI \cdot \cos \angle IAK$. Letting $X$ be the point of contact of the incircle of $ABC$ with side $AC$, we have $AX=MN$. Thus, $MN=\frac{117+120-125}{2}=\boxed{056}$

Solution 3 (Bash)

Project $I$ onto $AC$ and $BC$ as $D$ and $E$. $ID$ and $IE$ are both in-radii of $\triangle ABC$ so we get right triangles with legs $r$ (the in-radius length) and $s - c = 56$. Since $IC$ is the hypotenuse for the 4 triangles ($\triangle INC, \triangle IMC, \triangle IDC,$ and $\triangle IEC$), $C, D, M, I, N, E$ are con-cyclic on a circle we shall denote as $\omega$ which is also the circumcircle of $\triangle CMN$ and $\triangle CDE$. To find $MN$, we can use the Law of Cosines on $\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})$ where $O$ is the center of $\omega$. Now, the circumradius $R$ can be found with Pythagorean Theorem with $\triangle CDI$ or $\triangle CEI$: $r^2 + 56^2 = (2R)^2$. To find $r$, we can use the formula $rs = [ABC]$ and by Heron's, $[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}$. To find $\angle MCN$, we can find $\angle MIN$ since $\angle MCN = 180 - \angle MIN$. $\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}$. Thus, $\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}$ and since $\angle A + \angle B + \angle C = 180$, we have $\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}$. Plugging this into our Law of Cosines (LoC) formula gives $MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})$. To find $\cos{\angle C}$, we use LoC on $\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}$. Our formula now becomes $MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}$. After simplifying, we get $MN^2 = 3136 \implies MN = \boxed{056}$.

--lucasxia01

Solution 4

Because $\angle CMI = \angle CNI = 90$, $CMIN$ is cyclic.

Ptolemy on CMIN:

$CN*MI+CM*IN=CI*MN$

$CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN$

$MN = CI \sin \angle MCN$ by angle addition formula.

$\angle MCN = 180 - \angle MIN = 90 - \angle BCI$.

Let $H$ be where the incircle touches $BC$, then $CI \cos \angle BCI = CH = \frac{a+b-c}{2}$. $a=120, b=117, c=125$, for a final answer of $\boxed{056}$.

Video Solution

https://www.youtube.com/watch?v=yIUBhWiJ4Dk ~Mathematical Dexterity

Video Solution

https://www.youtube.com/watch?v=vkniYGN45F4

~Shreyas S

Alternate Solution: https://www.youtube.com/watch?v=L2OzYI0OJsc&t=12s

See also

2011 AIME I (Problems • Answer Key • Resources)
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Problem 3 		Followed by
Problem 5
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