Problem 1574

1. Assume the points on the sides:
Let $P$ be a point on $OA$ such that $OP = x$, $Q$ be a point on $OB$ such that $OQ = y$, and $R$ be a point on $OC$ such that $OR = z$.

2. Apply the Law of Cosines:
Since $\triangle PQR$ is an equilateral triangle, the lengths of $PQ$, $QR$, and $RP$ are equal. Using the Law of Cosines in $\triangle PQR$:
$$PQ^2 = x^2 + y^2 - xy \cos(\theta)$$
$$QR^2 = y^2 + z^2 - yz \cos(\theta)$$
$$RP^2 = z^2 + x^2 - zx \cos(\theta)$$
where $\theta$ is the angle between the sides of the tetrahedron.

3. Equating the sides:
Since $PQ = QR = RP$, we have:
$$x^2 + y^2 - xy = y^2 + z^2 - yz = z^2 + x^2 - zx$$

4. Simplify the equations:
From $x^2 + y^2 - xy = y^2 + z^2 - yz$, we get:
$$x^2 - xy = z^2 - yz$$
Rearrange to:
$$x^2 - z^2 = xy - yz$$
Factorize:
$$(x + z)(x - z) = y(x - z)$$
Since $x \neq z$ and $y \neq 0$, we can divide both sides by $x - z$:
$$x + z = y$$

5. Repeat for other pairs:
Similarly, from $y^2 + z^2 - yz = z^2 + x^2 - zx$, we get:
$$y^2 - yz = x^2 - zx$$
Rearrange to:
$$y^2 - x^2 = yz - zx$$
Factorize:
$$(y + x)(y - x) = z(y - x)$$
Since $y \neq x$ and $z \neq 0$, we can divide both sides by $y - x$:
$$y + x = z$$

6. Conclude the equality:
From $x + z = y$ and $y + x = z$, we can conclude that $x = y = z$.

7. Parallelism of sides:
Since $x = y = z$, the points $P, Q, R$ divide the sides $OA, OB, OC$ in the same ratio. By the properties of similar triangles, $\triangle PQR$ is similar to $\triangle ABC$, and the sides $PQ, QR, RP$ are parallel to $AB, BC, CA$ respectively.

$\blacksquare$