2019 AMC 12A Problems/Problem 19
Contents
Problem
In with integer side lengths,
What is the least possible perimeter for
?
Solution 1
We intend to use law of sines, so let's flip all the cosines (Sine is positive for , so we're good there).
These are in the ratio , so our minimal triangle has side lengths
,
, and
.
is our answer.
-Rowechen Zhong
Solution 2
is obtuse since its cosine is negative, so we let the foot of the altitude from
to
be
. Let
,
,
, and
. By the Pythagorean Theorem,
and
. Thus,
. The sides of the triangle are then
,
, and
, so for some integers
,
and
, where
and
are minimal. Hence,
, or
. Thus smallest possible positive integers
and
that satisfy this are
and
, so
. The sides of the triangle are
,
, and
, so
is our answer.
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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All AMC 12 Problems and Solutions |
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