Problem 1133

Since the roles of $A$ and $C$ can be interchanged in the problem, we can assume that the triangle $ABC$ has a positive orientation. Let $AB = c$, $BC = a$, and $\angle ABC = \beta$, and denote by $A'$ the point obtained by rotating $A$ around $B$ by $-90^\circ$ (see figure). Then $A'$ is also a lattice point, because if the coordinates of $\overrightarrow{BA}$ are $(x, y)$, then the coordinates of $\overrightarrow{BA'}$ are $(y, -x)$, which are integers (since both $A$ and $B$ are lattice points), and thus the coordinates of $A'$ are also integers, as they are sums of integers.

!

In the triangle $A'BC$, by the cosine rule,

$$A'C^2 = BC^2 + BA'^2 - 2 \cdot BC \cdot BA' \cdot \cos(90^\circ - \angle ABC) = a^2 + c^2 - 2ac \sin \beta$$

According to our condition, $(a + c)^2 < 8T + 1$, which, using the formula $T = \frac{1}{2} ac \sin \beta$, can be rewritten as

$$-2ac \sin \beta < \frac{1}{2} \left(1 - (a + c)^2\right)$$

Thus,

$$A'C^2 = a^2 + c^2 - 2ac \sin \beta < a^2 + c^2 + \frac{1}{2} \left(1 - (a + c)^2\right) = \frac{1}{2} \left(1 + (a - c)^2\right)$$

By the triangle inequality, $A'C \geq |BC - A'B| = |a - c|$ (equality is possible if the triangle $A'BC$ is degenerate), so $A'C^2 \geq (a - c)^2$. Therefore,

$$(a - c)^2 \leq A'C^2 < \frac{1}{2} \left(1 + (a - c)^2\right)$$

From this, we get $(a - c)^2 < 1$. Substituting this back into the inequality (1),

$$A'C^2 < \frac{1}{2} (1 + 1) = 1$$

follows. However, the distance between two lattice points can only be less than 1 if the points coincide, so $A' \equiv C$.

Thus, the point $A'$ obtained by rotating $A$ around $B$ by $-90^\circ$ is $C$, meaning that $A$, $B$, and $C$ are indeed three vertices of a square.

Terpai Tamás (Fazekas M. Fóv. Gyak. Gimn., 12th grade)