Difference between revisions of "2020 AIME I Problems/Problem 1"
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=== Solution 2 === | === Solution 2 === | ||
− | We have <math>\triangle BRQ\sim \triangle DRC</math> so <math>\frac{112}{RC} = \frac{BR}{DR}</math>. We also have <math>\triangle BRC \sim \triangle DRP</math> so <math>\frac{ RC}{847} = \frac {BR}{DR}</math>. Equating the two results gives <math>\frac{ | + | We have <math>\triangle BRQ\sim \triangle DRC</math> so <math>\frac{112}{RC} = \frac{BR}{DR}</math>. We also have <math>\triangle BRC \sim \triangle DRP</math> so <math>\frac{ RC}{847} = \frac {BR}{DR}</math>. Equating the two results gives <math>\frac{13}{RC} = \frac{ RC}{17797}</math> and so <math>RC^2=13*17797</math> which solves to <math>RC=\boxed{481}</math> |
Revision as of 16:58, 27 February 2020
Contents
Problem
Let be a parallelogram. Extend
through
to a point
and let
meet
at
and
at
Given that
and
find
Solution
Solution 1
There are several similar triangles. , so we can write the proportion:
Also, , so:
Substituting,
Thus, .
Solution 2
We have so
. We also have
so
. Equating the two results gives
and so
which solves to