2012 AMC 12B Problem 19

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?

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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.

Tagged: cube geometry · Pythagorean Theorem · symmetry