2012 AMC 12B Problem 19

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}

Answer: A

Concepts:cube geometry Pythagorean Theorem symmetry

Difficulty rating: 2110

Small hint:

By symmetry each octahedron vertex is the same distance

t

from

P_1

(or

P_1'

) along its edge

Big hint:

Set the squared distance between two vertices sharing

P_1

equal to the squared distance across, and solve for

t

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

Answer: A

Difficulty rating: 2110

Small hint:

Big hint:

Solution:

Problem 19 in other years