|
|
| Line 4: |
Line 4: |
| | | | |
| | == 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.
| |
| − | <center>
| |
| − | <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>
| |
| − | </center>
| |
| − | 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}} |
, 
, 
and 
. The angle bisector of angle 
intersects 
at point 
, and the angle bisector of angle 
intersects 
at point 
. Let 
and 
be the feet of the perpendiculars from 
to 
and 
, respectively. Find 
.
and 
such that they intersects lines 
at points 
and 
, respectively.
Since 
is the angle bisector of angle B,and 
, M is the midpoint of 
.For the same reason,
,N is the midpoint of 
.
Hence
PQ=BP+AQ-AB=120+117-125=112,so
.
