Problem 1609

1. Identify the given information and set up the problem:
- We have two triangles with the same perimeter.
- The first triangle is an equilateral triangle with integer side lengths.
- The second triangle has side lengths 1, $d$, and another side which we will determine.

2. Determine the side lengths of the equilateral triangle:
- Since the first triangle is equilateral, all its sides are equal.
- Let the side length of the equilateral triangle be $a$.
- Therefore, the perimeter of the equilateral triangle is $3a$.

3. Determine the perimeter of the second triangle:
- The second triangle has sides 1, $d$, and another side which we will call $x$.
- The perimeter of the second triangle is $1 + d + x$.

4. Set the perimeters equal to each other:
- Since both triangles have the same perimeter, we have:
$$3a = 1 + d + x$$

5. Apply the triangle inequality to the second triangle:
- For the second triangle to be valid, it must satisfy the triangle inequality:
$$1 + d > x, \quad 1 + x > d, \quad d + x > 1$$
- From $1 + d > x$, we get $x < 1 + d$.
- From $1 + x > d$, we get $x > d - 1$.
- From $d + x > 1$, we get $x > 1 - d$. Since $d$ is an integer and $d \geq 1$, this inequality is always satisfied.

6. **Determine the possible value of $x$:**
- Combining $x < 1 + d$ and $x > d - 1$, we get:
$$d - 1 < x < 1 + d$$
- Since $x$ must be an integer, the only possible value for $x$ that satisfies these inequalities is $x = d$.

7. **Substitute $x = d$ into the perimeter equation:**
- Substituting $x = d$ into $3a = 1 + d + x$, we get:
$$3a = 1 + d + d \implies 3a = 1 + 2d$$

8. **Solve for $d$ modulo 3:**
- Rearrange the equation to isolate $d$:
$$3a = 1 + 2d \implies 2d = 3a - 1$$
- Taking both sides modulo 3, we get:
$$2d \equiv 3a - 1 \pmod{3}$$
- Since $3a$ is a multiple of 3, we have $3a \equiv 0 \pmod{3}$:
$$2d \equiv -1 \pmod{3} \implies 2d \equiv 2 \pmod{3}$$
- Dividing both sides by 2 (noting that 2 has a multiplicative inverse of 2 modulo 3), we get:
$$d \equiv 1 \pmod{3}$$

$\blacksquare$

The final answer is $\boxed{ d \equiv 1 \pmod{3} }$