Putnam 1998, Problem A-6

Let \( A, B, C \) denote distinct points with integer coordinates in \( \mathbb{R}^2 \). Prove that if
\[
(|AB| + |BC|)^2 < 8 \cdot [ABC] + 1
\]
then \( A, B, C \) are three vertices of a square. Here \( |XY| \) is the length of segment \( XY \) and \( [ABC] \) is the area of triangle \( ABC \).

Solution:
Recall the inequalities \( |AB|^2 + |BC|^2 \geq 2|AB||BC| \) (AM-GM) and \( |AB||BC| \geq 2[ABC] \) (Law of Sines). Also recall that the area of a triangle with integer coordinates is half an integer (if its vertices lie at \( (0,0), (p,q), (r,s) \), the area is \( |ps - qr|/2 \)), and that if \( A \) and \( B \) have integer coordinates, then \( |AB|^2 \) is an integer (Pythagoras). Now observe that
\[
8[ABC] \leq |AB|^2 + |BC|^2 + 4[ABC] \leq |AB|^2 + |BC|^2 + 2|AB||BC| < 8[ABC] + 1,
\]
and that the first and second expressions are both integers. We conclude that \( 8[ABC] = |AB|^2 + |BC|^2 + 4[ABC] \), and so \( |AB|^2 + |BC|^2 = 2|AB||BC| = 4[ABC] \); that is, \( B \) is a right angle and \( AB = BC \), as desired.