Difference between revisions of "2016 AIME II Problems/Problem 5"
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− | Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ac}{b}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{dfrac{ac}{b}}{1-\dfrac{a}{c}}</math>. Testing triangles of the form <math>2p+1, 2p^{2}+2p, 2p^{2}+2p+1</math>, we have <math>13, 84, 85</math> satisfy this equation, so <math>p=13+84+85=\boxed{182}</math>. | + | Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ac}{b}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{\dfrac{ac}{b}}{1-\dfrac{a}{c}}</math>. Testing triangles of the form <math>2p+1, 2p^{2}+2p, 2p^{2}+2p+1</math>, we have <math>13, 84, 85</math> satisfy this equation, so <math>p=13+84+85=\boxed{182}</math>. |
Revision as of 20:17, 17 March 2016
Triangle has a right angle at
. Its side lengths are pariwise relatively prime positive integers, and its perimeter is
. Let
be the foot of the altitude to
, and for
, let
be the foot of the altitude to
in
. The sum
. Find
.
Solution
Note that by counting the area in 2 ways, the first altitude is . By similar triangles, the common ratio is
for reach height, so by the geometric series formula, we have
. Testing triangles of the form
, we have
satisfy this equation, so
.