Question

Three of the edges of a cube are $\overline{AB},$ $\overline{BC},$ and $\overline{CD},$ and $\overline{AD}$ is an interior diagonal. Points $P,$ $Q,$ and $R$ are on $\overline{AB},$ $\overline{BC},$ and $\overline{CD},$ respectively, so that $AP = 5,$ $PB = 15,$ $BQ = 15,$ and $CR = 10.$ What is the area of the polygon that is the intersection of plane $PQR$ and the cube?

答案

Solution:

The cube has side $20.$ Take $B = (0,0,0),$ $A = (20,0,0),$ $C = (0,20,0),$ and $D = (0,20,20),$ so $\overline{AD}$ is an interior diagonal. Then $P = (15, 0, 0),$ $Q = (0, 15, 0),$ $R = (0, 20, 10),$ and the plane through them is $2x + 2y - z = 30.$

Evaluating $2x + 2y - z$ at the cube's vertices and checking all twelve edges, the plane also crosses the edges at $(5, 20, 20),$ $(20, 5, 20),$ and $(20, 0, 10),$ so the cross-section is the hexagon with vertices $(15,0,0),$ $(0,15,0),$ $(0,20,10),$ $(5,20,20),$ $(20,5,20),$ $(20,0,10)$ in order. Its projection onto the $xy$ -plane is the hexagon $(15,0),$ $(0,15),$ $(0,20),$ $(5,20),$ $(20,5),$ $(20,0),$ whose area by the shoelace formula is $175.$

The plane's unit normal $\frac{1}{3}(2, 2, -1)$ has vertical component of magnitude $\frac{1}{3},$ so projecting onto the $xy$ -plane multiplies area by $\frac{1}{3}.$ The cross-section therefore has area $3 \cdot 175 = 525.$

1998 AIME Problem 11

Solution:

Problem 11 in Other Years