2014 AMC 10A Problem 19

The segment $XY$ goes down through the stack, and its height drops steadily from top to bottom. So the part inside the cube of edge $3$ is the same fraction of $XY$ as that cube's height is of the whole stack.

1. Heights of the cubes. Stack from the bottom: the cube of edge $4$ covers heights $0$ to $4$, the cube of edge $3$ covers $4$ to $7$, the cube of edge $2$ covers $7$ to $9$, and the cube of edge $1$ covers $9$ to $10$. The whole stack is $10$ tall.
2. Set up 3-D coordinates. Use $x$ for left-right, $y$ for front-back, and $z$ for height. In the picture every cube is pushed into the same front-left corner. $X$ is the top front-left corner of the small cube, so $X = (0, 0, 10)$. $Y$ is the bottom back-right corner of the big cube, so $Y = (4, 4, 0)$.
3. Full length. The 3-D distance formula adds the squares of the three changes:
$$XY = \sqrt{4^2 + 4^2 + 10^2} = \sqrt{132} = \sqrt{4 \cdot 33} = 2\sqrt{33}.$$
4. The fraction. Along $XY$, the height falls from $10$ to $0$ at a steady rate. The cube of edge $3$ covers heights $4$ to $7$, a stretch of $3$ out of $10$. So it holds $\frac{3}{10}$ of the segment.
5. Is that piece really inside the cube? At height $7$ the segment has gone $\frac{3}{10}$ of the way, so it is at $x = y = 1.2$. At height $4$ it has gone $\frac{6}{10}$ of the way, so $x = y = 2.4$. Both are inside the $3$ by $3$ base of that cube, so yes.
6. Length. $\frac{3}{10} \cdot 2\sqrt{33} = \frac{6\sqrt{33}}{10} = \frac{3\sqrt{33}}{5}$.

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

Flat horizontal layers cut a straight segment into pieces in the same ratio as the layers' thicknesses, just as parallel lines cut a slanted line in proportion. So the only real work is the full length of $XY$, and for that you must read the picture as a solid: $Y$ is to the right of, behind, and below $X$. With any 3-D picture, give each labelled point coordinates before measuring.

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