Problem:
Four cubes with edge lengths 1,2,3, and 4 are stacked as shown. What is the length of the portion of XY contained in the cube with edge length 3?
Answer Choices:
A. 5333
B. 23
C. 3233
D. 4
E. 32
Solution:
Label vertices A,B, and C as shown. Note that XC=10 and CY=42+42=42. Because △XYC is a right triangle, XY=102+(42)2=233. The ratio of BX to CX is 53, so in the top face of the bottom cube the distance from B to XY is 42⋅53=5122. This distance is less than 32, so XY pierces the top and bottom faces of the cube with side length 3. The ratio of AB to XC is 103, so the length of XY that is inside the cube with side length 3 is 103⋅233=(A)5333.
OR
Place the figure in a 3-dimensional coordinate system with the lower left front corner at (0,0,0),X=(0,0,10), and Y=(4,4,0). Then line XY consists of all points of the form (4t,4t,10−10t). This line intersects the bottom face of the cube with side length 3 when 10−10t=4, or t=53; this is the point (512,512,4), and because 512<3, the point indeed lies on that face. Similarly, line XY intersects the top face of the cube with side length 3 when 10−10t=7, or t=103; this is the point (56,56,7). Therefore the desired length is
