AMC 8 2012 Problem 23

AMC8 2012 年第 23 题

四边形与多边形★★★☆☆

An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?

一个等边三角形和一个正六边形的周长相等。若三角形的面积是 4，那么六边形的面积是多少？

A. $\hspace{.05in}4$
B. $\hspace{.05in}5$
C. $\hspace{.05in}6$
D. $\hspace{.05in}4\sqrt3$
E. $\hspace{.05in}6\sqrt3$

正确答案 · Correct Answer

解析 · Solution

Solution 1

Let the perimeter of the equilateral triangle be

3s

. The side length of the equilateral triangle would then be

s

and the sidelength of the hexagon would be

\displaystyle \frac{s}{2}

. A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio

1: 4

, since the sidelength of the small equilateral triangle is half the side length of the large one. Thus, the area of one of the small equilateral triangles is 1. The area of the hexagon is then 1×6 = 6. Let the side length of the small triangle be

2x

, so the side length of the hexagon is

x

. By the Pythagorean theorem, the height of the triangle is

x\sqrt{3}

, so the area of the triangle is

\displaystyle \frac{x^2}{\sqrt{3}}

. We are given that the area is 4, so x must be

\displaystyle \frac{2}{\sqrt[4]{3}}

. Therefore, by the Pythagorean theorem, the height of one of the triangles in the hexagon (made by subdividing the hexagon into 6 congruent triangles), is

\displaystyle \frac{\sqrt{3}}{\sqrt[4]{3}}

, so the area is

\displaystyle \frac{\frac{2}{\sqrt[4]{3}}\times\frac{\sqrt{3}}{\sqrt[4]{3}}}{2}

, which equals 1. Since there are 6 such triangles in the hexagon, the area of the hexagon is 6. The area of an equilateral triangle is

\displaystyle \frac{\sqrt{3}}{4} a^2

, where a is the side. Since a hexagon is six equilateral triangles combined, the area of a hexagon is

\displaystyle \left(\frac{3\sqrt{3}}{2}\right) a^2

. If we assume the side of the hexagon is 1, and the side of the triangle is 2 (perimeter is the same but sides are 6:3), then the area of the triangle will be

4\sqrt{3}

. We killed it is a

\sqrt{3}: 4

ratio of the assumed triangle area to the actual triangle area. This indicates the area of the hexagon will also be a ratio of

\sqrt{3}: 4

, so the area of the assumed hexagon is

\displaystyle \frac{3\sqrt{3}}{2}

. Therefore, the area of the actual hexagon will be

\displaystyle \frac{\frac{3\sqrt{3}}{2} \cdot 4}{\sqrt{3}}

. Simplifying the result will give us 6.

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