, 2025

Solution:
Assume for sake of contradiction that such a dissection exists. It has exactly half as many $30^{\circ}$ angles as $75^{\circ}$ angles.

Around any intersection point except the square's vertices, the only angles that can appear are $30^{\circ}$, $75^{\circ}$, and $180^{\circ}$. The only combinations of these that sum to $180^{\circ}$ or $360^{\circ}$ are
$$6 \cdot 30^{\circ} = 180^{\circ},$$
$$30^{\circ} + 2 \cdot 75^{\circ} = 180^{\circ},$$
$$180^{\circ} = 180^{\circ},$$
$$12 \cdot 30^{\circ} = 360^{\circ},$$
$$7 \cdot 30^{\circ} + 2 \cdot 75^{\circ} = 360^{\circ},$$
$$2 \cdot 30^{\circ} + 4 \cdot 75^{\circ} = 360^{\circ},$$
$$6 \cdot 30^{\circ} + 180^{\circ} = 360^{\circ},$$
$$30^{\circ} + 2 \cdot 75^{\circ} + 180^{\circ} = 360^{\circ},$$
$$180^{\circ} + 180^{\circ} = 360^{\circ}.$$
In particular, around any such point, there are at least half as many $30^{\circ}$ angles as $75^{\circ}$ angles.

However, the square's vertices must each be surrounded by three $30^{\circ}$ angles and zero $75^{\circ}$ angles, as there is no other way to get a sum of $90^{\circ}$. Thus the total number of $30^{\circ}$ angles in the dissection must be at least 12 more than half the number of $75^{\circ}$ angles, contradiction.

Thus no such dissection exists.

Solution 2:
Again assume for sake of contradiction that a dissection exists. Interpret the dissection as a graph $G$, where the vertices of the graph are the vertices of all the triangles, and edges connect each pair of consecutive vertices along a line segment.

Call a vertex flat if it is on either the boundary of the square (including its corners) or the interior of an edge of any triangle. Let $X$ be the number of flat vertices and $Y$ be the number of non-flat vertices in $G$. Let $E$ and $F$ be the number of edges and faces (triangles) in the dissection, respectively. Then $(X + Y) - E + F = 1$.

Observing the angle combinations in the first solution, we see that any non-flat vertex must have at least 6 incident edges, and any flat vertex must have at least 4. Thus $2E \geq 6Y + 4X$, so $E \geq 3Y + 2X$.

The sum of the angles of all $F$ triangles is $\pi F$. Around any non-flat vertex, such angles sum to $2\pi$. Around any flat vertex, the angles sum to $\pi$, with the exception of the four corners of the square, where they sum to $\pi/2$ instead. Thus
$$F\pi = (X - 4)\pi + 4(\pi/2) + Y(2\pi) = (X + 2Y - 2)\pi,$$
so $F = X + 2Y - 2$. This means
$$X + Y - E + F \leq (X + Y) - (3Y + 2X) + (X + 2Y - 2) = -2,$$
contradiction. Thus no dissection exists.