Bogdanov I.I.

A sphere inscribed in the pyramid $SABC$ touches the faces $SAB, SBC, SCA$ at points $D, E, F$ respectively.

Find all possible values of the sum of the angles $SDA, SEB$, and $SFC$.
Since triangles SCE and $S C F$ are equal, $\angle S F C=\angle S E C$. Similarly, $\angle S E B=\angle S D B$ and $\angle S D A=\angle S F A$. Therefore,

$\angle S D A+\angle S E B+\angle S F C=\angle S F A+\angle S D B+\angle S E C=1 / 2\left(3 \cdot 360^{\circ}-(\angle A D B+\angle B E C+\angle C F A)\right)$. But the angles $A D B, B E C$, $C F A$ are equal to the angles under which the sides of the face ABC are seen from the point of its tangency with the inscribed sphere, hence their sum is $360^{\circ}$.

## Answer

$360^{\circ}$.

Given a regular hexagon with center . Six equal circles with centers at the vertices of the hexagon are drawn such that the point is inside the circles. An angle of magnitude $\alpha$ with its vertex at the point cuts off six arcs on these circles. Prove that the total magnitude of these arcs is $6 \alpha$.

## Solution

Consider the angle symmetric to the given one relative to the point . By the theorem on the magnitude of the angle between chords, these two angles cut off two arcs of total magnitude $2 \alpha$ on each circle, and on all six circles, 12 angles of total magnitude $12 \alpha$. Since the picture is symmetric relative to , each of the angles cuts off six arcs of total magnitude $6 \alpha$.

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