AMC12 2024 B

AMC12 2024 B · Q23

AMC12 2024 B · Q23. It mainly tests Circle theorems, 3D geometry (volume).

A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V.$ Segments $\overline{AV}$ and $\overline{DV}$ are perpendicular. What is the square of the height of the pyramid?

一个直角金字塔以边长为$1$的正八边形$ABCDEFGH$为底面，顶点为$V$。线段$\overline{AV}$和$\overline{DV}$互相垂直。金字塔高度的平方是多少？

(A) 1 1

(B) \frac{1+\sqrt2}{2} \frac{1+\sqrt2}{2}

(D) \frac32 \frac32

(E) \frac{2+\sqrt2}{3} \qquad \frac{2+\sqrt2}{3} \qquad

Answer

Correct choice: (B)

正确答案：（B）

Solution

To find the height of the pyramid, we need the length from the center of the octagon (denote as $I$) to its vertices and the length of AV. From symmetry, we know that $\overline{AV} = \overline{DV}$, therefore $\triangle{AVD}$ is a 45-45-90 triangle. Denote $\overline{AV}$ as $x$ so that $\overline{AD} = x\sqrt{2}$. Doing some geometry on the isosceles trapezoid $ABCD$ (we know this from the fact that it is a regular octagon) reveals that $\overline{AD}=1+2(\sqrt{2}/2)=1+\sqrt{2}$ and $\overline{AV}=(\overline{AD})/\sqrt{2}=(\sqrt{2}+2)/2$. To find the length $\overline{IA}$, we cut the octagon into 8 triangles, each with a smallest angle of 45 degrees. Using the law of cosines on $\triangle{AIB}$ we find that ${\overline{IA}}^2=(2+\sqrt{2})/2$. Finally, using the pythagorean theorem, we can find that ${\overline{IV}}^2={\overline{AV}}^2-{\overline{IA}}^2= {((\sqrt{2}+2)/2)}^2 - (2+\sqrt{2})/2 = \boxed{(1+\sqrt{2})/2}$ which is answer choice $\boxed{B}$.

要找到金字塔的高度，我们需要八边形中心（记为$I$）到其顶点的距离以及$AV$的长度。由对称性，我们知道$\overline{AV} = \overline{DV}$，因此$\triangle{AVD}$是45-45-90三角形。设$\overline{AV}$为$x$，则$\overline{AD} = x\sqrt{2}$。在等腰梯形$ABCD$上做一些几何计算（我们知道这是正八边形），发现$\overline{AD}=1+2(\sqrt{2}/2)=1+\sqrt{2}$，且$\overline{AV}=(\overline{AD})/\sqrt{2}=(\sqrt{2}+2)/2$。要找到$\overline{IA}$的长度，我们将八边形分成8个三角形，每个最小角度为45度。在$\triangle{AIB}$上使用余弦定律，我们发现${\overline{IA}}^2=(2+\sqrt{2})/2$。最后，使用勾股定理，我们可以找到${\overline{IV}}^2={\overline{AV}}^2-{\overline{IA}}^2= {((\sqrt{2}+2)/2)}^2 - (2+\sqrt{2})/2 = \boxed{(1+\sqrt{2})/2}$，这是答案选项$\boxed{B}$。

Topics