Question

A unit cube has vertices $P_1,$ $P_2,$ $P_3,$ $P_4,$ $P_1',$ $P_2',$ $P_3',$ and $P_4'.$ Vertices $P_2,$ $P_3,$ and $P_4$ are adjacent to $P_1,$ and for $1 \le i \le 4,$ vertices $P_i$ and $P_i'$ are opposite to each other. A regular octahedron has one vertex in each of the segments $P_1P_2,$ $P_1P_3,$ $P_1P_4,$ $P_1'P_2',$ $P_1'P_3',$ and $P_1'P_4'.$ What is the octahedron's side length?

$\dfrac{3\sqrt{2}}{4}$

$\dfrac{7\sqrt{6}}{16}$

$\dfrac{\sqrt{5}}{2}$

$\dfrac{2\sqrt{3}}{3}$

$\dfrac{\sqrt{6}}{2}$

Solution:

Place $P_1$ at the origin with edges along the axes, and let each of the three octahedron vertices near $P_1$ be a distance $t$ from $P_1;$ by symmetry the three near $P_1'$ are also a distance $t$ from $P_1'.$

Two vertices sharing $P_1,$ such as $(t,0,0)$ and $(0,t,0),$ are a distance $t\sqrt2$ apart. A vertex near $P_1,$ say $(t,0,0),$ and the appropriate vertex near $P_1',$ say $(1,1-t,1),$ must be the same distance apart.

Setting the two squared side lengths equal and using the cube's unit edges yields $t=\tfrac34,$ so the side length is $t\sqrt2=\dfrac{3\sqrt2}{4}.$

Thus, the correct answer is A.

2012 AMC 12B Problem 19

Solution:

Problem 19 in Other Years