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 \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> | 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> | ||
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+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtube.com/watch?v=OEkt8HbUVbQ&t=737s |
Latest revision as of 22:05, 23 December 2022
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
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