1991 USAMO Problems/Problem 1
Problem
In triangle , angle
is twice angle
, angle
is obtuse, and the three side lengths
are integers. Determine, with proof, the minimum possible perimeter.
Solution
After drawing the triangle, also draw the angle bisector of , and let it intersect
at
. Notice that
, and let
. Now from similarity,
However, from the angle bisector theorem, we have
but
is isosceles, so
so all sets of side lengths which satisfy the conditions also meet the boxed condition.
Notice that or else we can form a triangle by dividing
by their greatest common divisor to get smaller integer side lengths, contradicting the perimeter minimality. Since
is squared,
must also be a square because if it isn't, then
must share a common factor with
, meaning it also shares a common factor with
, which means
share a common factor, contradiction. Thus we let
, so
, and we want the minimal pair
.
By the Law of Cosines,
Substituting yields
. Since
,
. For
there are no integer solutions. For
, we have
that works, so the side lengths are
and the minimal perimeter is
.
See also
1991 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |