2011 AIME I Problems/Problem 4

Problem 4

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 intersects lines ${AB}$ at points $P$ and $Q$ , respectively. Since ${BM}$ is the angle bisector of angle B,and ${CM}$ is perpendicular to ${BM}$ ,so , $BP=BC=120$ , M is the midpoint of ${CP}$ .For the same reason, $AQ=AC=117$ ,N is the midpoint of ${CQ}$ . Hence $MN=\frac{PQ}{2}$ .But $PQ=BP+AQ-AB=120+117-125=112$ ,so $MN=\boxed{56}$ .

Solution 2

Let $I$ be the incenter of $ABC$ . Now, since $AM \perp LC$ and $AN \perp KB$ , we have $AMIN$ is a cyclic quadrilateral. Consequently, $\frac{MN}{\sin \angle MIN} = 2R = AI$ . Since $\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK$ , 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 $MN=\frac{117+120-125}{2}=\boxed{56}$

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 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 $MN^2 = 3136 \implies MN = \boxed{056}$ . $\square$

--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.

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

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 $\boxed{056}$ .

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