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AMC10 2022 A

AMC10 2022 A · Q21

AMC10 2022 A · Q21. It mainly tests Area & perimeter, Geometry misc.

A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$. The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?
一个碗形是由一个边长为$1$的正方形附着四个边长为$1$的正六边形形成的。相邻六边形的边重合,如图所示。通过连接碗沿上边缘四个六边形的顶端八个顶点得到的八边形的面积是多少?
stem
(A) 6 6
(B) 7 7
(C) 5+2\sqrt{2} 5+2\sqrt{2}
(D) 8 8
(E) 9 9
Answer
Correct choice: (B)
正确答案:(B)
Solution
We extend line segments $\ell,m,$ and $n$ to their point of concurrency, as shown below: We claim that lines $\ell,m,$ and $n$ are concurrent: In the lateral faces of the bowl, we know that lines $\ell$ and $m$ must intersect, and lines $\ell$ and $n$ must intersect. In addition, line $\ell$ intersects the top plane of the bowl at exactly one point. Since lines $m$ and $n$ are both in the top plane of the bowl, we conclude that lines $\ell,m,$ and $n$ are concurrent. In the lateral faces of the bowl, the dashed red line segments create equilateral triangles. So, the dashed red line segments all have length $1.$ In the top plane of the bowl, we know that $\overleftrightarrow{m}\perp\overleftrightarrow{n}.$ So, the dashed red line segments create an isosceles triangle with leg-length $1.$ Note that octagon has four pairs of parallel sides, and the successive side-lengths are $1,\sqrt2,1,\sqrt2,1,\sqrt2,1,\sqrt2,$ as shown below: The area of the octagon is \[3^2-4\cdot\left(\frac12\cdot1^2\right)=\boxed{\textbf{(B) }7}.\]
我们将线段 $\ell,m,$ 和 $n$ 向它们的交汇点延伸,如下图所示: 我们断言直线 $\ell,m,$ 和 $n$ 是并发于一点的:在碗的侧面上,我们知道 $\ell$ 和 $m$ 必须相交,$\ell$ 和 $n$ 也必须相交。此外,直线 $\ell$ 与碗的顶平面恰好相交于一点。由于直线 $m$ 和 $n$ 都在碗的顶平面内,我们得出直线 $\ell,m,$ 和 $n$ 是并发于一点的。 在碗的侧面上,红色虚线段形成了等边三角形。因此,红色虚线段的长度均为 $1$。在碗的顶平面内,我们知道 $\overleftrightarrow{m}\perp\overleftrightarrow{n}$。因此,红色虚线段形成了一个腿长为 $1$ 的等腰三角形。 注意该八边形有四对平行边,连续的边长依次为 $1,\sqrt2,1,\sqrt2,1,\sqrt2,1,\sqrt2$,如下图所示: 八边形的面积为 \[3^2-4\cdot\left(\frac12\cdot1^2\right)=\boxed{\textbf{(B) }7}\].
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