2011 AIME I Problems/Problem 4

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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 $IM=CI\cdot \frac{IP}{AI}$. Using a similar method, we can find that $NI=CI\cdot\frac{IQ}{BI}$. Therefore, our Ptolemy's expression simplifies to \[MN=\frac{IP}{AI}\cdot CN+\frac{IQ}{BI}\cdot MC=r\left(\frac{CN}{AI}+\frac{MC}{BI}\right),\] where $r$ is the inradius of triangle $ABC$. Thus, $r=[ABC]/181$. Also, right triangles $CNA$ and $CBM$ tell us that $CN=117\sin \frac{A}{2}$ and $MC=120\sin\frac{B}{2}$. But then $[CLA]=\frac{117\cdot AL}{2}\sin\frac{A}{2}$, and this is equal to $\frac{117}{242}\cdot [ABC]$ by the Angle Bisector Theorem. Therefore, solving this for $\sin\frac{A}{2}$ and substituting yields $CN=\frac{2\cdot 117\cdot [ABC]}{242\cdot AL}$. Similarly, $MC=\frac{2\cdot 120\cdot [ABC]}{245\cdot BK}$. We now replace these in our Ptolemy's expression to get \[MN=\frac{[ABC]}{181}\left(\frac{2\cdot 117\cdot [ABC]}{242\cdot AL\cdot AI}+\frac{2\cdot 120\cdot [ABC]}{245\cdot BK\cdot BI}\right)\]\[=\frac{2[ABC]^2}{181}\left(\frac{117}{242\cdot AL\cdot AI}+\frac{120}{245\cdot BK\cdot BI}\right).\]


We can also use mass points, assigning masses of $120$, $117$, and $125$ to points $A$, $B$, and $C$, respectively. Then point $L$ has a mass of $242$ and point $K$ has a mass of $245$, so $AI=\frac{242}{362}\cdot AL$ and $BI=\frac{245}{362}\cdot BK$. This simplifies our expression further to \[MN=4[ABC]^2\left(\frac{117}{242^2\cdot AL^2}+\frac{120}{245^2\cdot BK^2}\right).\]

Then using the angle bisector formula, we find that $AL^2=125\cdot 117\left(1-\frac{120^2}{242^2}\right)$ and $BK^2=125\cdot 120\left(1-\frac{117^2}{245^2}\right)$. Also, Heron's Formula tells us that \[[ABC]^2=181\cdot 61\cdot 56\cdot 64,\] so when we substitute this all in, we get \[MN=4\cdot 181\cdot 61\cdot 56\cdot 64\cdot\left(\frac{1}{125(242^2-120^2)}+\frac{1}{125(245^2-117^2)}\right)\]\[=\frac{4\cdot 181\cdot 61\cdot 56\cdot 64}{125}\cdot\left(\frac{1}{122\cdot362}+\frac{1}{128\cdot362}\right)\]\[=\frac{2\cdot 61\cdot 56\cdot 64}{125}\cdot\left(\frac{250}{122\cdot 128}\right)=\boxed{56}.\]

See also

2011 AIME I (ProblemsAnswer KeyResources)
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

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