AMC12 2012 B

AMC12 2012 B · Q19

AMC12 2012 B · Q19. It mainly tests 3D geometry (volume), Geometry misc.

A unit cube has vertices $P_1,P_2,P_3,P_4,P_1',P_2',P_3',$ and $P_4'$. Vertices $P_2$, $P_3$, and $P_4$ are adjacent to $P_1$, and for $1\le i\le 4,$ vertices $P_i$ and $P_i'$ are opposite to each other. A regular octahedron has one vertex in each of the segments $P_1P_2$, $P_1P_3$, $P_1P_4$, $P_1'P_2'$, $P_1'P_3'$, and $P_1'P_4'$. What is the octahedron's side length?

一个单位立方体有顶点 $P_1,P_2,P_3,P_4,P_1',P_2',P_3',$ 和 $P_4'$。顶点 $P_2$, $P_3$, 和 $P_4$ 与 $P_1$ 相邻，且对于 $1\le i\le 4,$ 顶点 $P_i$ 和 $P_i'$ 互为对顶点。正八面体在每一段 $P_1P_2$, $P_1P_3$, $P_1P_4$, $P_1'P_2'$, $P_1'P_3'$, 和 $P_1'P_4'$ 上各有一个顶点。这个正八面体的边长是多少？

(A) \dfrac{3\sqrt{2}}{4} \dfrac{3\sqrt{2}}{4}

(B) \dfrac{7\sqrt{6}}{16} \dfrac{7\sqrt{6}}{16}

(D) \dfrac{2\sqrt{3}}{3} \dfrac{2\sqrt{3}}{3}

(E) \dfrac{\sqrt{6}}{2} \dfrac{\sqrt{6}}{2}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Error creating thumbnail: Unable to save thumbnail to destination Observe the diagram above. Each dot represents one of the six vertices of the regular octahedron. Three dots have been placed exactly x units from the $(0,0,0)$ corner of the unit cube. The other three dots have been placed exactly x units from the $(1,1,1)$ corner of the unit cube. A red square has been drawn connecting four of the dots to provide perspective regarding the shape of the octahedron. Observe that the three dots that are near $(0,0,0)$ are each $(x)(\sqrt{2}$) from each other. The same is true for the three dots that are near $(1,1,1).$ There is a unique $x$ for which the rectangle drawn in red becomes a square. This will occur when the distance from $(x,0,0)$ to $(1,1-x, 1)$ is $(x)(\sqrt{2}$). Using the distance formula we find the distance between the two points to be: $\sqrt{{(1-x)^2} + {(1-x)^2} + 1}$ = $\sqrt{2x^2 - 4x +3}$. Equating this to $(x)(\sqrt{2}$) and squaring both sides, we have the equation: $2{x^2} - 4x + 3$ = $2{x^2}$ $-4x + 3 = 0$ $x$ = $\frac{3} {4}$. Since the length of each side is $(x)(\sqrt{2}$), we have a final result of $\frac{3 \sqrt{2}}{4}$. Thus, Answer choice $\boxed{\text{A}}$ is correct. (If someone can draw a better diagram with the points labeled P1,P2, etc., I would appreciate it).

Error creating thumbnail: Unable to save thumbnail to destination 观察上图。每个点表示正八面体的 6 个顶点之一。有 3 个点放在距离单位立方体的角 $(0,0,0)$ 恰好为 $x$ 的位置。另 3 个点放在距离角 $(1,1,1)$ 恰好为 $x$ 的位置。用红色连结其中 4 个点画出一个矩形，以帮助观察八面体的形状。注意靠近 $(0,0,0)$ 的三个点两两之间距离都是 $(x)(\sqrt{2}$)。靠近 $(1,1,1)$ 的三个点也同样如此。存在唯一的 $x$ 使得红色所画的矩形变为正方形。这发生在从 $(x,0,0)$ 到 $(1,1-x, 1)$ 的距离等于 $(x)(\sqrt{2}$) 时。用距离公式可得两点距离为：$\sqrt{{(1-x)^2} + {(1-x)^2} + 1}$ = $\sqrt{2x^2 - 4x +3}$。令其等于 $(x)(\sqrt{2}$)，并两边平方，得到方程： $2{x^2} - 4x + 3$ = $2{x^2}$ $-4x + 3 = 0$ $x$ = $\frac{3} {4}$。由于每条边长为 $(x)(\sqrt{2}$)，最终结果为 $\frac{3 \sqrt{2}}{4}$。因此答案选项 $\boxed{\text{A}}$ 正确。 (If someone can draw a better diagram with the points labeled P1,P2, etc., I would appreciate it).

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