Difference between revisions of "2011 AIME I Problems/Problem 4"
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== Solution == | == Solution == | ||
+ | |||
+ | === Solution 1 === | ||
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. | 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. | ||
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Substituting the values of <math>ON, MQ, OQ</math>, we have that the answer is <math>58.5 + 60 - 62.5 = \boxed {56}</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>IM=CI\cdot \frac{IP}{AI}</math>. Using a similar method, we can find that <math>NI=CI\cdot\frac{IQ}{BI}</math>. Therefore, our Ptolemy's expression simplifies to <cmath>MN=\frac{IP}{AI}\cdot CN+\frac{IQ}{BI}\cdot MC=r\left(\frac{CN}{AI}+\frac{MC}{BI}\right),</cmath> | ||
+ | where <math>r</math> is the inradius of triangle <math>ABC</math>. Thus, <math>r=[ABC]/181</math>. Also, right triangles <math>CNA</math> and <math>CBM</math> tell us that <math>CN=117\sin \frac{A}{2}</math> and <math>MC=120\sin\frac{B}{2}</math>. But then <math>[CLA]=\frac{117\cdot AL}{2}\sin\frac{A}{2}</math>, and this is equal to <math>\frac{117}{242}\cdot [ABC]</math> by the Angle Bisector Theorem. Therefore, solving this for <math>\sin\frac{A}{2}</math> and substituting yields <math>CN=\frac{2\cdot 117\cdot [ABC]}{242\cdot AL}</math>. Similarly, <math>MC=\frac{2\cdot 120\cdot [ABC]}{245\cdot BK}</math>. We now replace these in our Ptolemy's expression to get | ||
+ | <cmath>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)</cmath><cmath>=\frac{2[ABC]^2}{181}\left(\frac{117}{242\cdot AL\cdot AI}+\frac{120}{245\cdot BK\cdot BI}\right).</cmath> | ||
+ | |||
+ | |||
+ | We can also use [[mass points]], assigning masses of <math>120</math>, <math>117</math>, and <math>125</math> to points <math>A</math>, <math>B</math>, and <math>C</math>, respectively. Then point <math>L</math> has a mass of <math>242</math> and point <math>K</math> has a mass of <math>245</math>, so <math>AI=\frac{242}{362}\cdot AL</math> and <math>BI=\frac{245}{362}\cdot BK</math>. This simplifies our expression further to | ||
+ | <cmath>MN=4[ABC]^2\left(\frac{117}{242^2\cdot AL^2}+\frac{120}{245^2\cdot BK^2}\right).</cmath> | ||
+ | |||
+ | Then using the angle bisector formula, we find that <math>AL^2=125\cdot 117\left(1-\frac{120^2}{242^2}\right)</math> and <math>BK^2=125\cdot 120\left(1-\frac{117^2}{245^2}\right)</math>. Also, Heron's Formula tells us that <cmath>[ABC]^2=181\cdot 61\cdot 56\cdot 64,</cmath> | ||
+ | so when we substitute this all in, we get <cmath>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)</cmath><cmath>=\frac{4\cdot 181\cdot 61\cdot 56\cdot 64}{125}\cdot\left(\frac{1}{122\cdot362}+\frac{1}{128\cdot362}\right)</cmath><cmath>=\frac{2\cdot 61\cdot 56\cdot 64}{125}\cdot\left(\frac{250}{122\cdot 128}\right)=\boxed{56}.</cmath> | ||
== 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 00:35, 15 July 2013
Problem 4
In triangle , , 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 .
Solution
Solution 1
Extend such that it intersects lines and at points and , respectively.
Lemma 1: are midpoints of and
Proof: Consider the reflection of the vertex over the line , and let this point be . Since , we have that is the image of after reflection over , and from the definition of reflection . Then it is easily seen that since is an angle bisector, that , so lies on . Similarly, if we define to be the reflection of over , then we find that lies on . Then we can now see that , with a homothety of ratio taking the first triangle to the second. Then this same homothety takes everything on the line to everything on the line . So since lie on , this homothety also takes to so they are midpoints, as desired.
Lemma 2: are isosceles triangles
Proof: To show that is isosceles, note that , with similarity ratio of . So it suffices to show that triangle is isosceles. But this follows quickly from Lemma 1, since is both an altitude and an angle bisector of . is isosceles by the same reasoning.
Since is a midline, it then follows that and . Since and are both isosceles, we have that and . Since is a midline, . We want to find , which is just .
Substituting the values of , we have that the answer is .
Solution 2
Let be the intersection of and , or rather the incenter of triangle . Noting that and are right, we conclude that is a cyclic quadrilateral, so by Ptolemy's Theorem, Now let and be inradii to and respectively in the following picture, which is not to scale.
We know that . In triangle , we have Therefore, , and . Thus . Using a similar method, we can find that . Therefore, our Ptolemy's expression simplifies to where is the inradius of triangle . Thus, . Also, right triangles and tell us that and . But then , and this is equal to by the Angle Bisector Theorem. Therefore, solving this for and substituting yields . Similarly, . We now replace these in our Ptolemy's expression to get
We can also use mass points, assigning masses of , , and to points , , and , respectively. Then point has a mass of and point has a mass of , so and . This simplifies our expression further to
Then using the angle bisector formula, we find that and . Also, Heron's Formula tells us that so when we substitute this all in, we get
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.