Maths Olympiad Prep

Track / Stage 5 / 116 of 400 #716 of 1964

Problem 716

AIME late
Geometry Difficulty 5.3 Find the answer

2. Let ABCDABCDA B C D A' B' C' D' be a cube with edge aa, where ABCDA B C D is one of the faces of the cube, and AA,BB,CCA A', B B', C C', and DDD D' are edges of the cube. Calculate (in terms of aa) the height of the pyramid ACBDA C B' D', dropped from the vertex DD.

A number or a short expression. Spacing and $ signs are ignored.

Official solution

Solution. Let's note that all the edges of the pyramid ACBDA C B' D are equal and have a length of b=a2b=a \sqrt{2}. This means that ACBDA C B^{\prime} D^{\prime} is obtained from the given cube when four equal triangular pyramids ACBB,ACDD,ABDAA C B^{\prime} B, A C D^{\prime} D, A B^{\prime} D^{\prime} A^{\prime}, and CBDCC B^{\prime} D^{\prime} C^{\prime}, each with a volume of a36\frac{a^{3}}{6}, are removed. Therefore, the volume VV of ACBDA C B^{\prime} D^{\prime} is a33\frac{a^{3}}{3}. On the other hand,

!
for the volume VV, we have V=b2h312V=\frac{b^{2} h \sqrt{3}}{12}, where hh is the desired height. By equating the two volumes and solving, we get h=2a33h=\frac{2 a \sqrt{3}}{3}.

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