10,11

In a triangular pyramid DABC, the sums of three plane angles at each of the vertices $D, A$, and $B$ are equal to $180^{\circ}, DC=15$, $\angle ACB=60^{\circ}$. Find the radius of the circumscribed sphere if the radius of the inscribed sphere is 3.
Consider the unfolding C1BC2AC2DC1 of tetrahedron ABCD onto the plane of triangle ABD (Fig.1), where points C1, C2, and C3 are the vertices of triangles with bases BD, AD, and AB, respectively. Since the sums of the three plane angles at each of the vertices A, B, and D of tetrahedron ABCD are each 180°, point A lies on segment C2C3, point B lies on segment C1C3, and point D lies on segment C1C2, and A, B, and D are the midpoints of these segments. Therefore, AB, BD, and AD are the midlines of triangle C1C2C3. This means that the opposite edges of tetrahedron ABCD are pairwise equal. Therefore, the faces of the tetrahedron are equal triangles (by three sides), i.e., the tetrahedron is equifacial. We will prove that the center of the circumscribed sphere of such a tetrahedron coincides with the center of the inscribed sphere. Let O be the center of the sphere of radius R circumscribed around the given tetrahedron (Fig.2).

Perpendiculars dropped from point O to the faces of the tetrahedron pass through the centers of the circumscribed circles of these faces, and since the tetrahedron is equifacial, all its faces are equal acute triangles.

Therefore, the radii of their circumscribed circles are equal, and the centers of these circles are located inside the faces. Denote them by R1. Then the distances from point O to the planes of the faces are equal to \(\sqrt{R^2 - R_1^2} = r\). This means that point O is equidistant from all the faces of the tetrahedron by the same distance r. Therefore, O is the center of the inscribed sphere, and r is the radius of this sphere. Since \(AB = CD = 15\) and \(\angle ACB = 60^\circ\), then

\[
R_1 = \frac{AB}{2 \sin 60^\circ} = \frac{15}{\sqrt{3}} = 5 \sqrt{3}
\]

and since \(r = 3\),

\[
R = \sqrt{R_1^2 + r^2} = \sqrt{75 + 9} = \sqrt{84} = 2 \sqrt{21}
\]

Answer

\(2 \sqrt{21}\).

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