Putnam 2010, Problem B-2

Given that \( A \), \( B \), and \( C \) are noncollinear points in the plane with integer coordinates such that the distances \( AB \), \( AC \), and \( BC \) are integers, what is the smallest possible value of \( AB \)?

Solution:
The smallest distance is 3, achieved by $A = (0,0)$, $B = (3,0)$, $C = (0,4)$. To check this, it suffices to check that $AB$ cannot equal 1 or 2. (It cannot equal 0 because if two of the points were to coincide, the three points would be collinear.)

The triangle inequality implies that $|AC - BC| \leq AB$, with equality if and only if $A, B, C$ are collinear. If $AB = 1$, we may assume without loss of generality that $A = (0,0)$, $B = (1,0)$. To avoid collinearity, we must have $AC = BC$, but this forces $C = (1/2, y)$ for some $y \in \mathbb{R}$, a contradiction. (One can also treat this case by scaling by a factor of 2 to reduce to the case $AB = 2$, treated in the next paragraph.)

If $AB = 2$, then we may assume without loss of generality that $A = (0,0)$, $B = (2,0)$. The triangle inequality implies $|AC - BC| \in \{0,1\}$. Also, for $C = (x,y)$, $AC^2 = x^2 + y^2$ and $BC^2 = (2-x)^2 + y^2$ have the same parity; it follows that $AC = BC$. Hence $C = (1,y)$ for some $y \in \mathbb{R}$, so $y^2$ and $y^2 + 1 = BC^2$ are consecutive perfect squares. This can only happen for $y = 0$, but then $A, B, C$ are collinear, a contradiction again.

**Remark.** Manjul Bhargava points out that more generally, a Heronian triangle (a triangle with integer sides and rational area) cannot have a side of length 1 or 2 (and again it is enough to treat the case of length 2). The original problem follows from this because a triangle whose vertices have integer coordinates has area equal to half an integer (by Pick's formula or the explicit formula for the area as a determinant).