Difference between revisions of "2003 AIME I Problems/Problem 15"
(solution (mass points) by me@home, Asymptote [myself]) |
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Having determined <math>CB/CF</math>, we reassign mass points to determine <math>FE/FD</math>. This setup involves <math>\tri CFD</math> and [[transversal]] <math>MEB</math>. For simplicity, put masses of <math>240,289</math> at <math>C,F</math>. To find the mass we should put at <math>D</math>, we compute <math>CM/MD</math>: applying the Angle Bisector Theorem again and using the fact <math>M</math> is a midpoint, we find | Having determined <math>CB/CF</math>, we reassign mass points to determine <math>FE/FD</math>. This setup involves <math>\tri CFD</math> and [[transversal]] <math>MEB</math>. For simplicity, put masses of <math>240,289</math> at <math>C,F</math>. To find the mass we should put at <math>D</math>, we compute <math>CM/MD</math>: applying the Angle Bisector Theorem again and using the fact <math>M</math> is a midpoint, we find | ||
<cmath> | <cmath> | ||
| − | \frac { | + | \frac {MD}{CM} = \frac {169\cdot\frac {260}{289} - 130}{130} = \frac {49}{289} |
</cmath> | </cmath> | ||
At this point we could find the mass at <math>D</math> but it's unnecessary. | At this point we could find the mass at <math>D</math> but it's unnecessary. | ||
Revision as of 17:11, 11 December 2012
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
![[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
![\[\frac {MD}{CM} = \frac {169\cdot\frac {260}{289} - 130}{130} = \frac {49}{289}\]](http://latex.artofproblemsolving.com/e/c/5/ec5c61a08d192fbfdce38843f06fdacc623a9b22.png)
At this point we could find the mass at but it's unnecessary.
![\[\frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\frac {C}{D} = \frac {289}{240}\frac {49}{289} = \boxed{\frac {49}{240}}\]](http://latex.artofproblemsolving.com/f/9/f/f9f0ce34f5f373bed95ef055ef69226929142dd1.png)
and the answer is 
.
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 | ||
