Difference between revisions of "1991 USAMO Problems/Problem 1"
(→Solution 1: I made the code for my diagram code easy to refactor. You're welcome, future editors.) |
(→Solution 1: adjusting the position of the label of point A) |
||
Line 19: | Line 19: | ||
C = (6/11, 8/11); | C = (6/11, 8/11); | ||
− | label("$\mathsf{A}$", A, | + | label("$\mathsf{A}$", A, W); |
label("$\mathsf{B}$", B, E); | label("$\mathsf{B}$", B, E); | ||
label("$\mathsf{C}$", C, N); | label("$\mathsf{C}$", C, N); |
Revision as of 19:07, 20 April 2023
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
Solution 1
(diagram by integralarefun)
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—a 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
.
Alternate Solution
In let
. From the law of sines, we have
Thus the ratio
We can simplify
Likewise,
Letting
, rewrite
We find that to satisfy the conditions for an obtuse triangle, and therefore
.
The rational number with minimum denominator (in order to minimize scaling to obtain integer solutions) above is
, which also has a denominator divisible by 2 (to take advantage of the coefficients of 2 and 4 in the ratio and further minimize scaling).
Inserting into the ratio, we find
. When scaled minimally to obtain integer side lengths, we find
and that the perimeter is
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1991 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.