2006
Problem - 3105
A plane passing through the vertex $A$ and the center of its inscribed sphere of a tetrahedron $ABCD$ intersects its edge $BC$ and $CD$ at point $E$ and $F$, as shown. If $AEF$ divides this tetrahedron into two equal volume parts: $A-BDEF$ and $A-CEF$, what is the relationship between these two parts' surface areas $S_1$ and $S_2$ where $S_1 = S_{A-BDEF}$ and $S_1=S_{A-CEF}$? $(A) S_1 < S_2\quad(B) S_1 > S_2\quad (C) S_1 = S_2 \quad(D) $ cannot determine
Let the center and radius of its inscribed sphere be $O$ and $r$ respectively, and connect $AO$, $BO$, $CO$, $DO$. Now we find tetrahedron $A-BEFD$ can be divided into $O-ABD$, $O-ABE$, $O-AFD$ and $O-BEDF$, therefore $$V_{A-BEFD} = V_{O-ABD}+V_{O-ABE}+V_{O-AFD}+V_{O-BEDF}$$
Meanwhile, we note that the height of all these objects equals $r$ and the sum of all their bases equals $S_1$. This means $$V_{A-BEFD} = \frac{1}{3}\cdot r \cdot S_1$$
Similarly, we have $$V_{A-CEF} = \frac{1}{3}\cdot r \cdot S_2$$
Setting them equal leads to $S_1 = S_2$.