Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 3"
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| − | By the Power of a Point Theorem on <math>B</math>, we have <math>BD | + | By the Power of a Point Theorem on <math>B</math>, we have <math>BD \times BA=BP \times BQ=2 \times 4=8</math>. By the Power of a Point on <math>C</math>, we have <math>CE \times CA=CQ \times CP=5 \times 7=35</math>. Dividing these two results yields <math>\frac{BD \times BA}{CE \times CA}=\frac{8}{35}</math>. We are given <math>BD=CE</math> and so <math>\frac{BD}{CE}=1</math>. Then the previous equation simplifies to <math>\frac{AB}{AC}=\frac{8}{35}</math>. Hence <math>m+n=8+35=\boxed{043}</math> |
Revision as of 10:49, 9 March 2014
Problem
In triangle 
Points 
and 
are located on 
such that 
The circumcircle of 
cuts 
at 
respectively. If 
then the ratio 
can be expressed in the form 
where 
are relatively prime positive integers. Find 
Solution
![[asy] unitsize(1cm); draw((-1,0)--(8,0)--(0,4)--cycle); draw((0,4)--(1,0)); draw((0,4)--(3,0)); label("$A$",(0,4),N); label("$B$",(-1,0),WSW); label("$C$",(8,0),ESE); label("$P$",(1,0),S); label("$Q$",(3,0),S); draw(circumcircle((0,4),(1,0),(3,0))); label("$D$",(-0.6,2),SW); label("$E$",(4.5,1.7),NE); [/asy]](http://latex.artofproblemsolving.com/0/7/d/07d435eb6a55b041b35df03b02b9cb0cd4e7571d.png)
By the Power of a Point Theorem on 
, we have 
. By the Power of a Point on 
, we have 
. Dividing these two results yields 
. We are given 
and so 
. Then the previous equation simplifies to 
. Hence 
