Difference between revisions of "2009 AIME I Problems/Problem 12"
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Since <math>AB=37,</math> this means that the ratio of the perimeter of <math>\triangle ABI</math> to <math>AB</math> is just <math>\frac{8}{3},</math> so our answer is | Since <math>AB=37,</math> this means that the ratio of the perimeter of <math>\triangle ABI</math> to <math>AB</math> is just <math>\frac{8}{3},</math> so our answer is | ||
<cmath>8+3=\boxed{11}.</cmath> | <cmath>8+3=\boxed{11}.</cmath> | ||
+ | |||
+ | == See also == | ||
{{AIME box|year=2009|n=I|num-b=11|num-a=13}} | {{AIME box|year=2009|n=I|num-b=11|num-a=13}} |
Revision as of 21:36, 24 November 2022
Problem
In right with hypotenuse
,
,
, and
is the altitude to
. Let
be the circle having
as a diameter. Let
be a point outside
such that
and
are both tangent to circle
. The ratio of the perimeter of
to the length
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution
Note that
. Thus,
. We also find that
and
. From here, we let
. Thus,
so
. Observe that
and
Thus,
However, we also know that
Thus, we get
Thus, the perimeter of
is
which gives
Since
this means that the ratio of the perimeter of
to
is just
so our answer is
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |