Solution

The shape of the "sunbeam" does not depend on the shape of the hole because the principle of a pinhole camera (camera obscura) is at work here, and the light source (the Sun) is extended.

The angular size of the Sun when observed from Earth is $\alpha \approx 0.5^\circ \approx 9 \cdot 10^{-3}$ rad. Each point of the hole in the curtain can be considered as a point aperture projecting an image of the solar disk onto the wall. The diameter of this disk on the wall is $D = \alpha L$, where $L$ is the distance from the curtain to the wall.

The total light spot on the wall is formed as a superposition of a multitude of such round disks, whose centers are shifted relative to each other within the dimensions of the hole itself, $d$.
The characteristic size of the spot on the wall can be estimated as the sum: $D_{\text{total}} \approx d + \alpha L$.

This leads to two limiting cases:

1. $d \ll \alpha L$ (the hole is small or the wall is far). In this case, the size of the hole itself is negligible compared to the projection of the Sun. The multiple solar disks merge into almost a single one, and we see a distinct round image of the Sun on the wall. The shape of the hole itself only creates a microscopic blur at the edges of the spot, invisible to the eye.
2. $d \ge \alpha L$ (the hole is large or the wall is close). In this case, the size of the Sun's projection is small compared to the size of the hole. Now the shape of the spot exactly repeats the shape of the hole (triangle, square), and the angular size of the Sun $\alpha$ only creates a small penumbra (blur) of width $\alpha L$ at the edges of this spot. It is in this case that the shape of the sunbeam will begin to depend on the shape of the hole.

Absolutely the same logic applies to a shard of a mirror: if the size of the mirror is $d \ll \alpha L$, the sunbeam will be round (image of the Sun). If you take a large mirror ($d \ge \alpha L$), the shape of the sunbeam will repeat the shape of the mirror.