AMC12 2011 A

AMC12 2011 A · Q15

AMC12 2011 A · Q15. It mainly tests Similarity, Coordinate geometry.

The circular base of a hemisphere of radius $2$ rests on the base of a square pyramid of height $6$. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?

半径为 $2$ 的半球的圆形底面放在一个高为 $6$ 的正方锥的底面上。该半球与锥体其余四个侧面都相切。求该锥体底面的边长。

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

(B) \frac{13}{3} \frac{13}{3}

(D) 6 6

(E) \frac{13}{2} \frac{13}{2}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $ABCDE$ be the pyramid with $ABCD$ as the square base. Let $O$ and $M$ be the center of square $ABCD$ and the midpoint of side $AB$ respectively. Lastly, let the hemisphere be tangent to the triangular face $ABE$ at $P$. Notice that $\triangle EOM$ has a right angle at $O$. Since the hemisphere is tangent to the triangular face $ABE$ at $P$, $\angle EPO$ is also $90^{\circ}$. Hence $\triangle EOM$ is similar to $\triangle EPO$. $\frac{OM}{2} = \frac{6}{EP}$ $OM = \frac{6}{EP} \times 2$ $OM = \frac{6}{\sqrt{6^2 - 2^2}} \times 2 = \frac{3\sqrt{2}}{2}$ The length of the square base is thus $2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}$

设金字塔为 $ABCDE$，其中 $ABCD$ 为正方形底面。设 $O$ 为正方形 $ABCD$ 的中心，$M$ 为边 $AB$ 的中点。最后，设半球与三角形侧面 $ABE$ 在 $P$ 点相切。注意到 $\triangle EOM$ 在 $O$ 处为直角。由于半球与三角形侧面 $ABE$ 在 $P$ 点相切，$\angle EPO$ 也为 $90^{\circ}$。因此 $\triangle EOM$ 与 $\triangle EPO$ 相似。 $\frac{OM}{2} = \frac{6}{EP}$ $OM = \frac{6}{EP} \times 2$ $OM = \frac{6}{\sqrt{6^2 - 2^2}} \times 2 = \frac{3\sqrt{2}}{2}$ 因此正方形底面的边长为 $2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}$.

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