Problem #511
Solution
The three vertices of a triangle of sides a, b, and c are lattice points, and
these points lie on a circle of radius R. Prove that abc ≥ 2R.
Solution: The Extended Law of Sines states that for any triangle 4ABC,
a
b
c
=
=
= 2R.
sin A
sin B
sin C
On the other hand, if K is the area of the triangle, then is K = 12 ab sin C.
abc
Substituting sin C = 2R/c shows that K =
.
4R
By Pick’s Formula, the area of a triangle is
1
B + I − 1,
2
where B is the number of lattice points on the boundary of the triangle and
I is the number of lattice points inside the triangle. We know that I ≥ 3
because the vertices are integer points, and we know that B ≥ 0. Therefore
K ≥ 1/2, and so
abc
≥ 1/2,
4R
and the result follows.
Note: Instead of Pick’s Formula, one can use the determinantal formula,
which states that the area of a triangle with vertices at (x1 , y1 ), (x2 , y2 ), (x3 , y3 )
is the absolute value of
x y 1
1 1 1 x2 y2 1
2 x3 y3 1
This is a half of a positive integer, so it must be at least 1/2.
Source: 1971 Putnam Exam.