Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2011 AIME I Problems/Problem 4

Revision as of 01:58, 15 July 2013 by Mathgeek2006 (talk | contribs) (→‎Solution 2)

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

Solution 1

Extend ${MN}$ such that it intersects lines ${AC}$ and ${BC}$ at points $O$ and $Q$, respectively.


Lemma 1: $O, Q$ are midpoints of $AC$ and $BC$

Proof: Consider the reflection of the vertex $C$ over the line $BM$, and let this point be $C_1$. Since $\angle{BMC} = 90^{\circ}$, we have that $C_1$ is the image of $C$ after reflection over $M$, and from the definition of reflection $\angle{MBC} = \angle{MBC_1}$. Then it is easily seen that since $BM$ is an angle bisector, that $\angle{MBC_1} = \angle{MBA}$, so $C_1$ lies on $AB$. Similarly, if we define $C_2$ to be the reflection of $C$ over $N$, then we find that $C_2$ lies on $AB$. Then we can now see that $\triangle{CMN} \sim \triangle{CC_1C_2}$, with a homothety of ratio $2$ taking the first triangle to the second. Then this same homothety takes everything on the line $MN$ to everything on the line $AB$. So since $O, Q$ lie on $MN$, this homothety also takes $O, Q$ to $A, B$ so they are midpoints, as desired. $\Box$

Lemma 2: $\triangle{MQC}, \triangle{NOC}$ are isosceles triangles

Proof: To show that $\triangle{MQC}$ is isosceles, note that $\triangle{MQC} \sim \triangle{C_1BC}$, with similarity ratio of $\frac{1}{2}$. So it suffices to show that triangle $\triangle{C_1BC}$ is isosceles. But this follows quickly from Lemma 1, since $BM$ is both an altitude and an angle bisector of $\angle{C_1BC}$. $\triangle{NOC}$ is isosceles by the same reasoning. $\Box$


Since ${OQ}$ is a midline, it then follows that ${OC} = 58.5$ and ${QC} = 60$. Since $\triangle MQC$ and $\triangle NOC$ are both isosceles, we have that $ON = OC = 58.5$ and $MQ = QC = 60$. Since $OQ$ is a midline, $OQ = 62.5$. We want to find $MN$, which is just $ON + MQ - OQ$.

Substituting the values of $ON, MQ, OQ$, we have that the answer is $58.5 + 60 - 62.5 = \boxed {56}$.

Solution 2

Let $I$ be the intersection of $AL$ and $BK$, or rather the incenter of triangle $ABC$. Noting that $\angle IMC$ and $\angle CNI$ are right, we conclude that $CNIM$ is a cyclic quadrilateral, so by Ptolemy's Theorem, \[CI\cdot MN=IM\cdot CN+NI\cdot MC.\] Now let $IP$ and $IQ$ be inradii to $AC$ and $BC$ respectively in the following picture, which is not to scale.

[asy] size(200); pair A=(0,0),C=(117,0), B=(100,85),I=incenter(A,B,C),K=extension(B,I,A,C),L=extension(A,I,B,C),M=foot(C,B,K),EN=foot(C,A,L); D(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle,black); draw(B--M--C--EN--A); draw(M--K^^EN--L); MP("M",M,WNW);MP("N",EN,NNW);MP("L",L,ENE);MP("K",K,S);MP("I",I,NW); markscalefactor=.75; draw(rightanglemark(C,M,I)^^rightanglemark(I,EN,C)); pair FAC=foot(I,A,C); pair FBC=foot(I,B,C); MP("P",FAC,S);MP("Q",FBC,ESE); draw(I--FAC^^I--FBC,dotted); draw(C--I); draw(rightanglemark(I,FAC,A)); dot(A);dot(B);dot(C);dot(M);dot(EN);dot(K);dot(L);dot(FAC);dot(FBC);dot(I); [/asy]

We know that $\frac{\angle A+\angle B+\angle C}{2}=90^\circ$. In triangle $CBM$, we have \[90^\circ=\angle CBM+\angle BCI+\angle ICM=\frac{\angle B}{2}+\frac{\angle C}{2}+\angle ICM.\] Therefore, $\angle ICM=\frac{\angle A}{2}$, and $\triangle CIM\sim\triangle AIP$. Thus $MC=CI\cdot \frac{AP}{AI}$. Using a similar method, we can find that $CN=CI\cdot \frac{BQ}{BI}$. Therefore, our Ptolemy's expression simplifies to \[MN=IM\cdot \frac{BQ}{BI}+NI\cdot\frac{AP}{AI}=IM\cdot\cos \frac{B}{2}+NI\cdot\cos\frac{A}{2}.\] Using right triangles $CIM$ and $NCI$, we also know that $IM=CI\cdot\sin \frac{A}{2}$ and $NI=CI\cdot\sin\frac{B}{2}$. Thus \[MN=CI\cdot\cos \frac{B}{2}\sin\frac{A}{2}+CI\cdot\cos\frac{A}{2}\sin\frac{B}{2}=CI\cdot\sin\left(\frac{A+B}{2}\right)=CI\cdot \cos \frac{C}{2}.\] But this last expression is equal to $CQ$. This a tangent to the incircle, so it has length $s-c=181-125=\boxed{56}$.

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&oldid=56650"