2020 AIME I Problems/Problem 1: Difference between revisions
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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 15:58, 27 February 2020
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 
