2015 AIME I Problems/Problem 11
Problem
Triangle has positive integer side lengths with
. Let
be the intersection of the bisectors of
and
. Suppose
. Find the smallest possible perimeter of
.
Solution 1 (No Trig)
Let and the foot of the altitude from
to
be point
and
. Since ABC is isosceles,
is on
. By Pythagorean Theorem,
. Let
and
. By angle bisector theorem,
. Also,
. Solving for
, we get
. Then, using Pythagorean Theorem on
we have
. Simplifying, we have
. Factoring out the
, we have
. Adding 1 to the fraction and simplifying, we have
. Crossing out the
, and solving for
yields
. Then, we continue as Solution 2 does.
Solution 2 (Trig)
Let be the midpoint of
. Then by SAS Congruence,
, so
.
Now let ,
, and
.
Then
and .
Cross-multiplying yields .
Since ,
must be positive, so
.
Additionally, since has hypotenuse
of length
,
.
Therefore, given that is an integer, the only possible values for
are
,
,
, and
.
However, only one of these values, , yields an integral value for
, so we conclude that
and
.
Thus the perimeter of must be
.
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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