AMC12 2008 A

AMC12 2008 A · Q18

AMC12 2008 A · Q18. It mainly tests Coordinate geometry, 3D geometry (volume).

Triangle $ABC$, with sides of length $5$, $6$, and $7$, has one vertex on the positive $x$-axis, one on the positive $y$-axis, and one on the positive $z$-axis. Let $O$ be the origin. What is the volume of tetrahedron $OABC$?

三角形 $ABC$ 的边长分别为 $5$、$6$ 和 $7$，其中一个顶点在正 $x$ 轴上，一个在正 $y$ 轴上，一个在正 $z$ 轴上。设 $O$ 为原点。四面体 $OABC$ 的体积是多少？

(A) \sqrt{85} \sqrt{85}

(B) \sqrt{90} \sqrt{90}

(D) 10 10

(E) \sqrt{105} \sqrt{105}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem, \begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*} so $a^2 = (5^2+7^2-6^2)/2 = 19$; similarly, $b^2 = 6$ and $c^2 = 30$. Since $OA$, $OB$, and $OC$ are mutually perpendicular, the tetrahedron's volume is \[abc/6\] because we can consider the tetrahedron to be a right triangular pyramid. \[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\] which is answer choice $\boxed{\text{C}}$.

不妨设 $A$ 在 $x$ 轴上，$B$ 在 $y$ 轴上，$C$ 在 $z$ 轴上，并令 $AB, BC, CA$ 的长度分别为 $5, 6, 7$。设线段 $OA,OB,OC$ 的长度分别为 $a,b,c$。由勾股定理， \begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*} 因此 $a^2 = (5^2+7^2-6^2)/2 = 19$；同理 $b^2 = 6$，$c^2 = 30$。由于 $OA$、$OB$、$OC$ 两两互相垂直，该四面体的体积为 \[abc/6\]，因为可将其视为直三棱锥。 \[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\] 对应选项为 $\boxed{\text{C}}$。

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