Track / Stage 5 / 120 of 400 #720 of 1964

Problem 720

AIME late

Geometry Difficulty 5.3 Find the answer

6. Given that $P, Q, R, S$ are four points inside the tetrahedron $A-BCD$ , and $Q, R, S, P$ are the midpoints of $PA, QB, RC, SD$ respectively. If $V_{P-ABC}$ represents the volume of the tetrahedron $P-ABC$ , and similarly for the others, then $V_{P-ABC}: V_{P-BCD}: V_{P-DAB}: V_{P-DAC}=$ $\qquad$ .

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

6. $8: 1: 4: 2$

Points $A, P$ to the distance of plane $BCD$ are $h_{A}, h_{P}$ , and similarly for the others. Therefore, we have
$\begin{array}{l} h_{Q}=\frac{1}{2}\left(h_{A}+h_{P}\right) \\ h_{R}=\frac{1}{2} h_{Q} \\ =\frac{1}{4}\left(h_{A}+h_{P}\right) \\ h_{S}=\frac{1}{2} h_{R}=\frac{1}{8}\left(h_{A}+h_{P}\right), \\ h_{P}=\frac{1}{2} h_{S}=\frac{1}{16}\left(h_{A}+h_{P}\right) . \end{array}$

Thus, $h_{P}=\frac{1}{15} h_{A}$ .
Hence $V_{P-BCD}: V_{A-BCD}=1: 15$ .
Similarly, $V_{P-CAD}: V_{A-BCD}=2: 15$ .
$\begin{array}{l} V_{P-DAB}: V_{A-BCD}=4: 15, \quad V_{P-ABC}: V_{A-BCD}=8: 15 \\ V_{P-ABC}: V_{P-BCD}: V_{P-DAB}: V_{P-DAC}=8: 1: 4: 2 \end{array}$

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