AMC10 2013 B

AMC10 2013 B · Q15

AMC10 2013 B · Q15. It mainly tests Area & perimeter, Polygons.

A wire is cut into two pieces, one of length $a$ and the other of length $b$. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $\frac{a}{b}$?

一根铁丝被剪成两段，一段长 $a$，另一段长 $b$。长 $a$ 的段弯成一个正三角形，长 $b$ 的段弯成一个正六边形。三角形和六边形的面积相等。$\frac{a}{b}$ 的值是多少？

(A) 1 1

(B) $\frac{\sqrt{6}}{2}$ $\frac{\sqrt{6}}{2}$

(D) 2 2

(E) $\frac{3\sqrt{2}}{2}$ $\frac{3\sqrt{2}}{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let $s$ be the side length of the triangle and $h$ the side length of the hexagon. The hexagon can be subdivided into 6 equilateral triangles by drawing segments from the center of the hexagon to each vertex. Because the areas of the large triangle and hexagon are equal, the triangles in the hexagon each have area $\frac{1}{6}$ of the area of the large triangle. Thus $\frac{h}{s}=\sqrt{\frac{1}{6}}$ so $h=\frac{\sqrt{6}}{6}s.$ The perimeter of the triangle is $a=3s$ and the perimeter of the hexagon is $b=6h=\sqrt{6}s$, so $\frac{a}{b}=\frac{3s}{\sqrt{6}s}=\frac{\sqrt{6}}{2}.$

答案（B）：设 $s$ 为三角形的边长，$h$ 为六边形的边长。通过从六边形中心向各顶点作线段，可将六边形分成 $6$ 个正三角形。由于大三角形与六边形的面积相等，六边形中的每个小三角形面积都是大三角形面积的 $\frac{1}{6}$。因此 $\frac{h}{s}=\sqrt{\frac{1}{6}}$，所以 $h=\frac{\sqrt{6}}{6}s。$ 三角形的周长为 $a=3s$，六边形的周长为 $b=6h=\sqrt{6}s$，因此 $\frac{a}{b}=\frac{3s}{\sqrt{6}s}=\frac{\sqrt{6}}{2}。$

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