Problem 1681

1. Given that the inradius $r$ of triangle $ABC$ is 1 and the side lengths $a, b, c$ are all integers, we need to prove that triangle $ABC$ is right-angled.

2. The area $S$ of the triangle can be expressed using the inradius and the semiperimeter $p$ as:
$$S = pr$$
Since $r = 1$, we have:
$$S = p$$
where $p = \frac{a + b + c}{2}$.

3. Using Heron's formula, the area $S$ of the triangle can also be written as:
$$S = \sqrt{p(p-a)(p-b)(p-c)}$$
Since $S = p$, we have:
$$p = \sqrt{p(p-a)(p-b)(p-c)}$$
Squaring both sides, we get:
$$p^2 = p(p-a)(p-b)(p-c)$$
Dividing both sides by $p$ (assuming $p \neq 0$):
$$p = (p-a)(p-b)(p-c)$$

4. Let us introduce new variables:
$$x = p - a, \quad y = p - b, \quad z = p - c$$
Then, we have:
$$a = p - x, \quad b = p - y, \quad c = p - z$$
Substituting these into the equation $p = (p-a)(p-b)(p-c)$, we get:
$$p = xyz$$

5. Since $x, y, z$ are positive integers and $x + y + z = p$, we have:
$$x + y + z = xyz$$
Without loss of generality, assume $x \leq y \leq z$. Then:
$$xyz = x + y + z \leq 3z$$
This implies:
$$xy \leq 3$$

6. The possible integer pairs $(x, y)$ that satisfy $xy \leq 3$ and $x, y \geq 1$ are:
$$(1, 1), (1, 2), (1, 3), (2, 1), (3, 1)$$

7. We need to check which of these pairs can form a valid triangle with integer sides. For each pair, we calculate $z$ and check if $x + y + z = xyz$.

- For $(x, y) = (1, 2)$:
$$z = 3 \quad \text{(since \( x + y + z = 1 + 2 + 3 = 6 \) and \( xyz = 1 \cdot 2 \cdot 3 = 6 \))}$$
Thus, $x = 1, y = 2, z = 3$.

8. The sides of the triangle are:
$$a = p - x = 3 - 1 = 2, \quad b = p - y = 3 - 2 = 1, \quad c = p - z = 3 - 3 = 0$$
This does not form a valid triangle.

9. For $(x, y) = (1, 3)$:
$$z = 2 \quad \text{(since \( x + y + z = 1 + 3 + 2 = 6 \) and \( xyz = 1 \cdot 3 \cdot 2 = 6 \))}$$
Thus, $x = 1, y = 3, z = 2$.

10. The sides of the triangle are:
$$a = p - x = 3 - 1 = 2, \quad b = p - y = 3 - 3 = 0, \quad c = p - z = 3 - 2 = 1$$
This does not form a valid triangle.

11. For $(x, y) = (2, 1)$:
$$z = 3 \quad \text{(since \( x + y + z = 2 + 1 + 3 = 6 \) and \( xyz = 2 \cdot 1 \cdot 3 = 6 \))}$$
Thus, $x = 2, y = 1, z = 3$.

12. The sides of the triangle are:
$$a = p - x = 3 - 2 = 1, \quad b = p - y = 3 - 1 = 2, \quad c = p - z = 3 - 3 = 0$$
This does not form a valid triangle.

13. For $(x, y) = (1, 1)$:
$$z = 4 \quad \text{(since \( x + y + z = 1 + 1 + 4 = 6 \) and \( xyz = 1 \cdot 1 \cdot 4 = 4 \))}$$
Thus, $x = 1, y = 1, z = 4$.

14. The sides of the triangle are:
$$a = p - x = 3 - 1 = 2, \quad b = p - y = 3 - 1 = 2, \quad c = p - z = 3 - 4 = -1$$
This does not form a valid triangle.

15. For $(x, y) = (3, 1)$:
$$z = 2 \quad \text{(since \( x + y + z = 3 + 1 + 2 = 6 \) and \( xyz = 3 \cdot 1 \cdot 2 = 6 \))}$$
Thus, $x = 3, y = 1, z = 2$.

16. The sides of the triangle are:
$$a = p - x = 3 - 3 = 0, \quad b = p - y = 3 - 1 = 2, \quad c = p - z = 3 - 2 = 1$$
This does not form a valid triangle.

17. The only valid set of sides that satisfy the conditions is $(3, 4, 5)$, which forms a right-angled triangle since:
$$3^2 + 4^2 = 5^2$$

Therefore, triangle $ABC$ is right-angled.

$\blacksquare$