Question

Distinct planes $p_1,$ $p_2,$ $\ldots,$ $p_k$ intersect the interior of a cube $Q.$ Let $S$ be the union of the faces of $Q$ and let $P = \bigcup_{j=1}^{k} p_j.$ The intersection of $P$ and $S$ consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of $Q.$ What is the difference between the maximum and the minimum possible values of $k?$

$8$

$12$

$20$

$23$

$24$

Answer: C

Concepts:cube geometry 3D geometry casework

Difficulty rating: 2460

Solution:

On every face, the required segments join midpoints of edges. A plane cutting the cube meets the faces in one of four symmetric shapes: a square through midpoints ( $3$ such planes), a rectangle per edge ( $12$ planes), a triangle per vertex ( $8$ planes), or a regular hexagon per pair of opposite vertices ( $4$ planes).

Using all of them gives the maximum $k = 3 + 12 + 8 + 4 = 27.$

The full figure consists of $24$ short segments and $12$ long segments. A square plane contains $(0,4),$ a rectangle $(2,2),$ a triangle $(3,0),$ and a hexagon $(6,0)$ short and long segments, respectively. Give each short segment weight $1$ and each long segment weight $3/2.$ Every plane then covers weight at most $6,$ whereas the required segments have total weight $24+(3/2)12=42.$ Thus at least $7$ planes are needed. The $4$ hexagon planes cover all $24$ short segments, and the $3$ square planes cover all $12$ long segments, so $k=7$ is attainable.

The difference is $27 - 7 = 20.$

Thus, the correct answer is C.

2012 AMC 12A Problem 22

Answer: C

Difficulty rating: 2460

Solution:

Problem 22 in Other Years