Difference between revisions of "2003 AIME I Problems/Problem 15"
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== Solution == | == Solution == | ||
+ | |||
+ | === Solution 1 === | ||
<center><asy> | <center><asy> | ||
size(400); pointpen = black; pathpen = black+linewidth(0.7); | size(400); pointpen = black; pathpen = black+linewidth(0.7); | ||
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</cmath> | </cmath> | ||
and the answer is <math>49 + 240 = \boxed{289}</math>. | and the answer is <math>49 + 240 = \boxed{289}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | By the Angle Bisector Theorem, we know that <math>[CBD]=\frac{169}{289}[ABC]</math>. Therefore, by finding the area of triangle <math>CBD</math>, we see that <cmath>\frac{507\cdot BD}{2}\sin\frac{B}{2}=\frac{169}{289}[ABC].</cmath> | ||
+ | Solving for <math>BD</math> yields <cmath>BD=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}}.</cmath> | ||
+ | Furthermore, <math>\cos\frac{B}{2}=\frac{BD}{BF}</math>, so <cmath>BF=\frac{BD}{\cos\frac{B}{2}}=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}\cos\frac{B}{2}}.</cmath> | ||
+ | Now by the identity <math>2\sin\frac{B}{2}\cos\frac{B}{2}=\sin B</math>, we get <cmath>BF=\frac{4[ABC]}{3\cdot289\sin B}.</cmath> | ||
+ | But then <math>[ABC]=\frac{360\cdot 507}{2}\sin B</math>, so <math>BF=\frac{240}{289}\cdot 507</math>. Thus <math>BF:FC=240:49</math>. | ||
+ | |||
+ | Now by the Angle Bisector Theorem, <math>CD=\frac{169}{289}\cdot 780</math>, and we know that <math>MC=\frac{1}{2}\cdot 780</math> so <math>DM:MC=\frac{169}{289}-\frac{1}{2}:\frac{1}{2}=49:289</math>. | ||
+ | |||
+ | We can now use mass points on triangle CBD. Assign a mass of <math>240\cdot 49</math> to point <math>C</math>. Then <math>D</math> must have mass <math>240\cdot 289</math> and <math>B</math> must have mass <math>49\cdot 49</math>. This gives <math>F</math> a mass of <math>240\cdot 49+49\cdot 49=289\cdot 49</math>. Therefore, <math>DE:EF=\frac{289\cdot 49}{240\cdot 289}=\frac{49}{240}</math>, giving us an answer of <math>\boxed{289}.</math> | ||
== See also == | == See also == |
Revision as of 22:39, 14 July 2013
Problem
In and
Let
be the midpoint of
and let
be the point on
such that
bisects angle
Let
be the point on
such that
Suppose that
meets
at
The ratio
can be written in the form
where
and
are relatively prime positive integers. Find
Solution
Solution 1
![[asy] size(400); pointpen = black; pathpen = black+linewidth(0.7); pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F); /* scale down by 100x */ D(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle); D(B--D(MP("D",D))--D(MP("F",F,NE))); D(B--D(MP("M",M))); MP("E",E,NE); D(rightanglemark(F,D,B,4)); MP("390",(M+C)/2); MP("390",(M+C)/2); MP("360",(A+B)/2,NW); MP("507",(B+C)/2,NE); [/asy]](http://latex.artofproblemsolving.com/b/0/7/b07153fb983b9e2b2209286c26983fb7db68ddb8.png)
For computation, instead consider the triangle as above except . In the following, let the name of a point represent the mass located there.
By the Angle Bisector Theorem, we can place mass points on of
respectively. Thus, a mass of
belongs at
(seen by reflecting
across
, to an image which lies on
).
Having determined
, we reassign mass points to determine
. This setup involves $\tri CFD$ (Error compiling LaTeX. Unknown error_msg) and transversal
. For simplicity, put masses of
at
. To find the mass we should put at
, we compute
: applying the Angle Bisector Theorem again and using the fact
is a midpoint, we find
At this point we could find the mass at
but it's unnecessary.
and the answer is
.
Solution 2
By the Angle Bisector Theorem, we know that . Therefore, by finding the area of triangle
, we see that
Solving for
yields
Furthermore,
, so
Now by the identity
, we get
But then
, so
. Thus
.
Now by the Angle Bisector Theorem, , and we know that
so
.
We can now use mass points on triangle CBD. Assign a mass of to point
. Then
must have mass
and
must have mass
. This gives
a mass of
. Therefore,
, giving us an answer of
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
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.