AMC10 2014 A

AMC10 2014 A · Q19

AMC10 2014 A · Q19. It mainly tests Coordinate geometry, 3D geometry (volume).

Four cubes with edge lengths 1, 2, 3, and 4 are stacked as shown. What is the length of the portion of $\overline{XY}$ contained in the cube with edge length 3?

四个边长分别为 1、2、3 和 4 的立方体按图堆叠。立方体边长为 3 的部分中 $\overline{XY}$ 包含的部分的长度是多少？

(A) $\frac{3\sqrt{33}}{5}$ $\frac{3\sqrt{33}}{5}$

(B) $2\sqrt{3}$ $2\sqrt{3}$

(D) 4 4

(E) $3\sqrt{2}$ $3\sqrt{2}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Label vertices $A, B,$ and $C$ as shown. Note that $XC = 10$ and $CY = \sqrt{4^2 + 4^2} = 4\sqrt{2}$. Because $\triangle XYC$ is a right triangle, $XY = \sqrt{10^2 + (4\sqrt{2})^2} = 2\sqrt{33}$. The ratio of $BX$ to $CX$ is $\frac{3}{5}$, so in the top face of the bottom cube the distance from $B$ to $XY$ is $4\sqrt{2} \cdot \frac{3}{5} = \frac{12\sqrt{2}}{5}$. This distance is less than $3\sqrt{2}$, so $XY$ pierces the top and bottom faces of the cube with side length 3. The ratio of $AB$ to $XC$ is $\frac{3}{10}$, so the length of $XY$ that is inside the cube with side length 3 is $\frac{3}{10} \cdot 2\sqrt{33} = \frac{3\sqrt{33}}{5}$.

按图标记顶点 $A, B$ 和 $C$。注意 $XC = 10$，$CY = \sqrt{4^2 + 4^2} = 4\sqrt{2}$。因为 $\triangle XYC$ 是直角三角形，故 $XY = \sqrt{10^2 + (4\sqrt{2})^2} = 2\sqrt{33}$。$BX$ 与 $CX$ 的比为 $\frac{3}{5}$，故在底层立方体顶面上，从 $B$ 到 $XY$ 的距离为 $4\sqrt{2} \cdot \frac{3}{5} = \frac{12\sqrt{2}}{5}$。此距离小于 $3\sqrt{2}$，故 $XY$ 穿过边长为 3 的立方体的顶面和底面。$AB$ 与 $XC$ 的比为 $\frac{3}{10}$，故边长为 3 的立方体内 $XY$ 的长度为 $\frac{3}{10} \cdot 2\sqrt{33} = \frac{3\sqrt{33}}{5}$。

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