In a triangular pyramid PABC, the sums of the three plane angles at each of the vertices A and B are both $180^{\circ}$ and $PC = AB$. Inside the pyramid, a point $D$ is taken, the sum of the distances from which to the three lateral faces PAB, PAC, and PBC is 7. Find the distance from the center of the circumscribed sphere to the face PAB, if the volumes of the pyramids PABC and BABC are in the ratio of $8: 1$.

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Consider the unfolding P1AP2CP3BP1 of the tetrahedron ABCD onto the plane of triangle ABC (Fig.1), where points P1, P2, and P3 are the vertices of triangles with bases AB, AC, and BC, respectively. Since the sums of the three plane angles at each of the vertices A and B of the tetrahedron PABC are $180^{\circ}$, point A lies on the segment P1P2, and point B lies on the segment P1P3, and A and B are the midpoints of these segments. Therefore, AB is the midline of triangle P1P2P3. This means that $P2P3 = 2AB$, and since $PC = AB$, then $CP2 = CP3 = AB$, so $CP2 + CP3 = P2P3$. This implies that point C lies on the segment P2P3, and C is the midpoint of this segment (Fig.2). Thus, AC, AB, and BC are the midlines of triangle P1P2P3. Therefore,

$$
BC = AP1 = AP, \quad AC = BP1 = BP
$$

i.e., the opposite edges of the tetrahedron ABCD are pairwise equal. This means that all faces of the tetrahedron are equal triangles (by three sides). We will prove that the center of the inscribed sphere of this 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.3). 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-angled 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 $\sqrt{R^2 - R_1} = r$. This means that point O is equidistant from all the faces of the tetrahedron at a distance of $r$. Therefore, O is the center of the inscribed sphere, and $r$ is the radius of this sphere. Let S be the area of each of the four faces of the tetrahedron PABC, H be the height of the tetrahedron drawn from vertex P, and x, y, z, and t be the heights of the pyramids DABP, DACP, DBCP, and DABC, respectively, drawn from their common vertex D (Fig.4). From the problem statement, it follows that $H = 8t$, and since

$$
V_{DABP} + V_{DACP} + V_{DBCP} + V_{DABC} = V_{PABC}
$$

we have

$$
\begin{gathered}
\frac{1}{3} S \cdot x + \frac{1}{3} S \cdot y + \frac{1}{3} S \cdot y + \frac{1}{3} S \cdot z = \\
= \frac{1}{3} S \cdot (x + y + z + t) = \frac{1}{3} S \cdot (7 + t) = \frac{1}{3} S \cdot H
\end{gathered}
$$

from which

$$
7 + t = H = 8t, \quad t = 1, \quad H = 8.
$$

Thus,

$$
V_{PABC} = \frac{1}{3} S \cdot H = \frac{1}{3} \cdot 8S.
$$

and since $V_{PABC} = \frac{1}{3} \cdot 4S \cdot r$, then

$$
r = \frac{3 V_{PABC}}{4S} = 2
$$

## Answer

2.00