Difference between revisions of "2003 AIME I Problems/Problem 15"
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== Solution == | == Solution == | ||
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+ | === 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
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. ! Undefined control sequence.) 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 |
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