AMC12 2021 B

AMC12 2021 B · Q14

AMC12 2021 B · Q14. It mainly tests Linear equations, Pythagorean theorem.

Let $ABCD$ be a rectangle and let $\overline{DM}$ be a segment perpendicular to the plane of $ABCD$. Suppose that $\overline{DM}$ has integer length, and the lengths of $\overline{MA},\overline{MC},$ and $\overline{MB}$ are consecutive odd positive integers (in this order). What is the volume of pyramid $MABCD?$

设 $ABCD$ 为矩形，$\overline{DM}$ 是垂直于 $ABCD$ 平面的线段。假设 $\overline{DM}$ 长度为整数，且 $\overline{MA},\overline{MC},$ 和 $\overline{MB}$ 的长度依次为连续的奇正整数（按此顺序）。金字塔 $MABCD$ 的体积是多少？

(A) 24\sqrt5 24\sqrt5

(B) 60 60

(D) 66 66

(E) 8\sqrt{70} 8\sqrt{70}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $MA=a$ and $MD=d.$ It follows that $MC=a+2$ and $MB=a+4.$ As shown below, note that $\triangle MAD$ and $\triangle MBC$ are both right triangles. By the Pythagorean Theorem, we have \begin{alignat*}{6} AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\ BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2. \end{alignat*} Since $AD=BC$ in rectangle $ABCD,$ we equate the expressions for $AD^2$ and $BC^2,$ then rearrange and factor: \begin{align*} a^2 - d^2 &= (a+4)^2 - (a+2)^2 \\ a^2 - d^2 &= 4a + 12 \\ a^2 - 4a - d^2 &= 12 \\ (a-2)^2 - d^2 &= 16 \\ (a+d-2)(a-d-2) &= 16. \end{align*} As $a+d-2$ and $a-d-2$ have the same parity, we get $a+d-2=8$ and $a-d-2=2,$ from which $(a,d)=(7,3).$ Applying the Pythagorean Theorem to right $\triangle MAD$ and right $\triangle MCD,$ we obtain $AD=2\sqrt{10}$ and $CD=6\sqrt2,$ respectively. Let the brackets denote areas. Together, the volume of pyramid $MABCD$ is \[\frac13\cdot [ABCD]\cdot MD = \frac13\cdot (AD\cdot CD)\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.\]

令 $MA=a$ 和 $MD=d$。则 $MC=a+2$ 和 $MB=a+4$。如下所示，注意 $\triangle MAD$ 和 $\triangle MBC$ 均为直角三角形。由勾股定理，有 \begin{alignat*}{6} AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\ BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2. \end{alignat*} 由于矩形 $ABCD$ 中 $AD=BC$，我们等式两边的表达式，然后重排并因式分解： \begin{align*} a^2 - d^2 &= (a+4)^2 - (a+2)^2 \\ a^2 - d^2 &= 4a + 12 \\ a^2 - 4a - d^2 &= 12 \\ (a-2)^2 - d^2 &= 16 \\ (a+d-2)(a-d-2) &= 16. \end{align*} 由于 $a+d-2$ 和 $a-d-2$ 同奇偶，我们取 $a+d-2=8$ 和 $a-d-2=2$，得 $(a,d)=(7,3)$。对直角 $\triangle MAD$ 和直角 $\triangle MCD$ 应用勾股定理，得 $AD=2\sqrt{10}$ 和 $CD=6\sqrt2$。令括号表示面积。金字塔 $MABCD$ 的体积为 \[\frac13\cdot [ABCD]\cdot MD = \frac13\cdot (AD\cdot CD)\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.\]

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