Track / Stage 6 / 121 of 400 #1121 of 2000

Problem 1121

National olympiad, first round

Geometry Difficulty 6.1 Prove it

15.10. (NPR, 76). A plane intersects three edges of a tetrahedron emanating from one vertex. Prove that this plane divides the surface of the tetrahedron into parts proportional to the volumes of the corresponding parts of the tetrahedron if and only if it passes through the center of the sphere inscribed in the tetrahedron.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

15.10. Let $V, S$ and $r$ denote the volume, surface area of a tetrahedron, and the radius of the inscribed sphere, respectively. One of the parts into which a plane divides the tetrahedron is a pyramid with its base lying in this plane. Let $V_{1}, S_{i}$ and $r_{\mathbf{i}}$ denote the volume, the lateral surface area of this pyramid, and the radius of the sphere with its center on the base of the pyramid, touching its lateral faces. The base of the pyramid passes through the center of the sphere inscribed in the tetrahedron if and only if $r=\boldsymbol{r}_{\mathbf{1}}$ . The latter equality, according to the formulas

$V=(1 / 3) S r_{,}, V_{i}=(1 / 3) S_{1} r_{1},$

is equivalent to the equality

$\frac{V}{V_{i}}=\frac{S}{S_{i}}, \text { i.e. } \frac{V-V_{i}}{V_{1}}=\frac{S-S_{i}}{S_{1}}$

which proves the statement of the problem.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.