Problem:
Three of the edges of a cube are AB,BC, and CD, and AD is an interior diagonal. Points P,Q, and R are on AB,BC, and 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?
Solution:
Because BP=BQ,PQ​ is parallel to AC. Thus the line through R that is parallel to PQ will intersect AF at U so that AU=CR. Because the line through R that is parallel to PQ​ is in plane PQR,U is a vertex of the intersection polygon. The midpoint of RU is also the center of the cube, so the intersection polygon has point symmetry with respect to the center of the cube. Hence its area is twice the area of isosceles trapezoid PQRU, whose altitude is
Set up a coordinate system so that A=(0,0,0),B=(20,0,0),C=(20,20,0), and D=(20,20,20). It follows that P=(5,0,0),Q=(20,15,0), and R=(20,20,10). Plane PQR can be described by an equation ax+by+cz=d. Substitute the coordinates of P,Q, and R into this equation to find that
d=5a=20a+15b=20a+20b+10c
hence that a=−b=2c. Thus plane PQR is described by 2x−2y+z=10. To find coordinates for the other points where the plane intersects the edges of the cube, replace two of the unknowns by 0 or 20, and solve for the third, which must also be between 0 and 20. This yields the three additional points S=(15,20,20), T=(0,5,20), and U=(0,0,10). The area of hexagon PQRSTU may now be found as above.
