2014 AMC 10A Problem 19

Set up coordinates with $x$ to the right, $y$ into the page, and $z$ upward. In the figure every cube shares the same left face ($x = 0$) and the same front face ($y = 0$); they differ only in height. Stacking from the bottom, the cubes occupy
$$4:\ 0 \le z \le 4,\qquad 3:\ 4 \le z \le 7,\qquad 2:\ 7 \le z \le 9,\qquad 1:\ 9 \le z \le 10.$$

$X$ is the top front-left corner of the smallest cube, $X = (0, 0, 10)$. $Y$ is the bottom back-right corner of the largest cube, $Y = (4, 4, 0)$. So
$$XY = \sqrt{4^2 + 4^2 + 10^2} = \sqrt{132} = 2\sqrt{33}.$$

Moving along $\overline{XY}$, the height drops at a constant rate from $10$ to $0$. The horizontal planes $z = 7$ and $z = 4$ therefore cut off exactly $\frac{7 - 4}{10} = \frac{3}{10}$ of the segment. (These points, $(1.2, 1.2, 7)$ and $(2.4, 2.4, 4)$, lie within the $3 \times 3$ footprint of the middle cube, so that piece really is inside the cube of edge $3$.)

The required length is
$$\frac{3}{10} \cdot 2\sqrt{33} = \frac{3\sqrt{33}}{5}.$$

The answer is $\boxed{\textbf{(A)}\ \dfrac{3\sqrt{33}}5}$.

A segment crossing a stack of horizontal slabs is divided in the same ratio as the heights of the slabs (parallel planes cut a transversal proportionally, the 3-D version of similar triangles). So the only real computation is the full length of $XY$, and for that the figure must be read as a solid: $Y$ is displaced from $X$ in all three directions. Whenever a diagram is drawn in oblique perspective, translate every labeled point into $(x, y, z)$ before measuring anything.

• Working in the flat side view with Y at the front-bottom corner, which gives length 2 root 29 and an answer not among the choices.
• Using the wrong height band for the edge-3 cube, for example $3 \le z \le 6$ from the top, which still gives $\frac{3}{10}$ here but fails in general; locate each cube from the bottom.
• Computing the fraction from horizontal displacement (edge $3$ out of $4$) instead of from height.