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

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== Solution ==  
== Solution ==  
Extend <math>{CM}</math> and <math>{CN}</math> such that they intersects lines <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively.  
=== Solution 1 ===
Since <math>{BM}</math> is the angle bisector of angle B,and <math>{CM}</math> is perpendicular to <math>{BM}</math> ,so , <math>BP=BC=120</math>, M is the midpoint of <math>{CP}</math> .For the same reason,<math>AQ=AC=117</math>,N is the midpoint of <math>{CQ}</math>.
Extend <math>{MN}</math> such that it intersects lines <math>{AC}</math> and <math>{BC}</math> at points <math>O</math> and <math>Q</math>, respectively.  
Hence<math>MN=\frac{PQ}{2}.But </math>PQ=BP+AQ-AB=120+117-125=112,so<math>MN=\boxed{56}</math>.
'''Lemma 1: <math>O, Q</math> are midpoints of <math>AC</math> and <math>BC</math>'''
'''Proof:''' Consider the reflection of the vertex <math>C</math> over the line <math>BM</math>, and let this point be <math>C_1</math>. Since <math>\angle{BMC} = 90^{\circ}</math>, we have that <math>C_1</math> is the image of <math>C</math> after reflection over <math>M</math>, and from the definition of reflection <math>\angle{MBC} = \angle{MBC_1}</math>. Then it is easily seen that since <math>BM</math> is an angle bisector, that <math>\angle{MBC_1} = \angle{MBA}</math>, so <math>C_1</math> lies on <math>AB</math>. Similarly, if we define <math>C_2</math> to be the reflection of <math>C</math> over <math>N</math>, then we find that <math>C_2</math> lies on <math>AB</math>. Then we can now see that <math>\triangle{CMN} \sim \triangle{CC_1C_2}</math>, with a homothety of ratio <math>2</math> taking the first triangle to the second. Then this same homothety takes everything on the line <math>MN</math> to everything on the line <math>AB</math>. So since <math>O, Q</math> lie on <math>MN</math>, this homothety also takes <math>O, Q</math> to <math>A, B</math> so they are midpoints, as desired. <math>\Box</math>
'''Lemma 2: <math>\triangle{MQC}, \triangle{NOC}</math> are isosceles triangles'''
'''Proof:''' To show that <math>\triangle{MQC}</math> is isosceles, note that <math>\triangle{MQC} \sim \triangle{C_1BC}</math>, with similarity ratio of <math>\frac{1}{2}</math>. So it suffices to show that triangle <math>\triangle{C_1BC}</math> is isosceles. But this follows quickly from Lemma 1, since <math>BM</math> is both an altitude and an angle bisector of <math>\angle{C_1BC}</math>. <math>\triangle{NOC}</math> is isosceles by the same reasoning. <math>\Box</math>
Since <math>{OQ}</math> is a midline, it then follows that <math>{OC} = 58.5</math> and <math>{QC} = 60</math>. Since <math>\triangle MQC</math> and <math>\triangle NOC</math> are both isosceles, we have that <math>ON = OC = 58.5</math> and <math>MQ = QC = 60</math>. Since <math>OQ</math> is a midline, <math>OQ = 62.5</math>. We want to find <math>MN</math>, which is just <math>ON + MQ - OQ</math>.
Substituting the values of <math>ON, MQ, OQ</math>, we have that the answer is <math>58.5 + 60 - 62.5 = \boxed {56}</math>.
=== Solution 2 ===
Let <math>I</math> be the intersection of <math>AL</math> and <math>BK</math>, or rather the incenter of triangle <math>ABC</math>. Noting that <math>\angle IMC</math> and <math>\angle CNI</math> are right, we conclude that <math>CNIM</math> is a cyclic quadrilateral, so by [[Ptolemy's Theorem]], <cmath>CI\cdot MN=IM\cdot CN+NI\cdot MC.</cmath>
Now let <math>IP</math> and <math>IQ</math> be inradii to <math>AC</math> and <math>BC</math> respectively in the following picture, which is not to scale.
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);
pair FAC=foot(I,A,C);
pair FBC=foot(I,B,C);
We know that <math>\frac{\angle A+\angle B+\angle C}{2}=90^\circ</math>. In triangle <math>CBM</math>, we have <cmath>90^\circ=\angle CBM+\angle BCI+\angle ICM=\frac{\angle B}{2}+\frac{\angle C}{2}+\angle ICM.</cmath> Therefore, <math>\angle ICM=\frac{\angle A}{2}</math>, and <math>\triangle CIM\sim\triangle AIP</math>.  
Thus <math>MC=CI\cdot \frac{AP}{AI}</math>. Using a similar method, we can find that <math>CN=CI\cdot \frac{BQ}{BI}</math>. Therefore, our Ptolemy's expression simplifies to <cmath>MN=IM\cdot \frac{BQ}{BI}+NI\cdot\frac{AP}{AI}=IM\cdot\cos  \frac{B}{2}+NI\cdot\cos\frac{A}{2}.</cmath>
Using right triangles <math>CIM</math> and <math>NCI</math>, we also know that <math>IM=CI\cdot\sin \frac{A}{2}</math> and <math>NI=CI\cdot\sin\frac{B}{2}</math>. Thus <cmath>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}.</cmath>
But this last expression is equal to <math>CQ</math>. This a tangent to the incircle, so it has length <math>s-c=181-125=\boxed{56}</math>.
== See also ==
== See also ==
{{AIME box|year=2011|n=I|num-b=3|num-a=5}}
{{AIME box|year=2011|n=I|num-b=3|num-a=5}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 03:27, 7 November 2015

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 $MN$.


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$MN=\boxed{56}$.

See also

2011 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AIME Problems and Solutions

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