Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 3"
(Created page with "==Problem== In triangle <math>ABC,</math> <math>BC=9.</math> Points <math>P</math> and <math>Q</math> are located on <math>BC</math> such that <math>BP=PQ=2,</math> <math>QC=5.<...") |
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==Solution== | ==Solution== | ||
− | By the Power of a Point Theorem on <math>B</math>, we have <math>BD*BA=BP*BQ=2*4=8</math>. By the Power of a Point on <math>C</math>, we have <math>CE*CA=CQ*CP=5*7=35</math>. Dividing these two results yields <math>\frac{BD*BA}{CE*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{ | + | ==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> | ||
+ | |||
+ | By the Power of a Point Theorem on <math>B</math>, we have <math>BD*BA=BP*BQ=2*4=8</math>. By the Power of a Point on <math>C</math>, we have <math>CE*CA=CQ*CP=5*7=35</math>. Dividing these two results yields <math>\frac{BD*BA}{CE*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 22:05, 1 January 2012
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
Solution
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