1. Given a triangular pyramid $S A B C$, the base of which is an equilateral triangle $A B C$, and all plane angles at vertex $S$ are equal to $\alpha$. For what least $\alpha$ can we assert that this pyramid is regular?

M. Malkin
Answer: $60^{\circ}$.

We will prove that when $\alpha=60^{\circ}$, the pyramid is regular. Let the sides of the triangle $ABC$ be equal to 1. Note that in any triangle with an angle of $60^{\circ}$, the side opposite this angle is the middle-length side (and if it is strictly less than one of the sides, it is strictly greater than the other). Suppose one of the lateral edges of the pyramid is not equal to 1, for example, $SA>1$. Then in the faces $SAB$ and $SAC$, the edges $SB$ and $SC$ will be less than 1, and thus in the face $SBC$, the edge $BC$ will not be the middle side, which is a contradiction.

Now we will show how to construct an irregular pyramid with dihedral angles $\alpha<60^{\circ}$ at the vertex $S$.

First method. First, we remove the lateral edge $SA$, and rotate the remaining lateral face $SBC$ around its base edge $BC$ until this face lies in the plane of the base and contains the base triangle. During this motion, there will be "current" pyramids, in which two equal dihedral angles are initially equal to $\alpha$, then greater than $\alpha$ (at the moment when the vertex projects onto the vertex of the base - since at this moment the sine of the angle at the vertex $S$ in the lateral faces $SAC$ and $SAB$ is $1: SB$, and in the face $SBC$ it is the ratio of the lateral height of this triangle to $SB$, which is less than 1), and at the end, in the "degenerate" pyramid, they are equal to $\alpha / 2$. Therefore, by continuity, there will be $\alpha$ again.

Second method. Consider the triangle $SAB$ with $SA=SB$ and $\angle S=\alpha$. Since $AB<SB$, there exists a point $C$ on the side $SA$ such that $BC=AB$. Now take a trihedral angle in which all dihedral angles are equal to $\alpha$, and mark segments $SA, SB, SC$ on its edges. The pyramid $SABC$ is the desired one.

Third method. (Sergey Komarov, 11th grade, MSU SUNT) Let $S'ABC$ be a regular pyramid with a dihedral angle $\alpha$ at the vertex $S'$. Let $X, Y$ be points such that $XY=AB$, and construct triangles $XYZ$ and $XYT$ on $XY$ such that $\angle ZXY=\angle ZYX=90^{\circ}-\frac{\alpha}{2}$, $\angle TXY=90^{\circ}+\frac{\alpha}{2}$, $\angle TXY=90^{\circ}-\frac{3\alpha}{2}$, then it is clear that $\angle XTY=\angle XZY=\alpha$. By the Law of Sines for these triangles, we have

$$
\frac{TY}{\sin \left(90^{\circ}+\frac{\alpha}{2}\right)}=\frac{XY}{\sin \alpha}=\frac{ZX}{\sin \left(90^{\circ}-\frac{\alpha}{2}\right)}=\frac{ZY}{\sin \left(90^{\circ}-\frac{\alpha}{2}\right)}
$$

which implies $TY=ZX=ZY$, since the sines of adjacent angles are equal.

Now let $S$ be a point on the extension of the segment $S'B$ beyond point $B$ such that $SB=TX$. Then $\angle SBC=\angle SBA=90^{\circ}+\frac{\alpha}{2}=\angle TXY, BC=BA=XY$, and $SB=SB=TX$, so $\triangle SBC=\triangle SBA=\triangle TXY$, from which $SC=SA=TY=ZX=ZY$, and thus $\triangle SCA=\triangle ZXY$ (by three sides). From all the indicated triangle equalities, it follows that the pyramid $SABC$ has all dihedral angles at the vertex $S$ equal to $\alpha$, so it satisfies the conditions of the problem. But it is not regular, since the line $S'B$ is not perpendicular to $ABC$, and on it only one point projects to the center of the triangle $ABC$, which is the point $S'$, and not the point $S$.