2013年AMC10竞赛B第15题
2013年$$AMC10$$竞赛$$B$$第$$15$$题  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 $$\frac ab$$?  一根电线被切成两块，一块长度为 $$a$$，另一块长度为 $$b$$。 长度$$a$$ 弯曲形成等边三角形，长度$$b$$ 弯曲形成正六边形。 三角形和六边形的面积相等。 求 $$\frac ab$$？
课内体系->知识点->解三角形->三角形面积公式 美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Regular Polygon
Using the area formulas for an equilateral triangle $$\left(\frac{s^{2}\sqrt{3}}{4}\right)$$ and regular hexagon $$\left(\frac{3s^{2}\sqrt{3}}{2}\right)$$ , with side length $$s$$ and plugging $$\frac a3$$ and $$\frac b6$$ into each equation, we find that $$\frac{a^{2}\sqrt{3}}{36}= \frac{b^{2}\sqrt{3}}{24}$$. Simplifying this, we get $$\frac ab=\boxed{\rm(B)\textasciitilde\frac{\sqrt{6}}{2}}$$.  The regular hexagon can be broken into $$6$$ small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle\textquotesingle s area is $$6$$ times the area of one of the little triangles. Therefore each side of the big triangle is $$\sqrt{6}$$ times the side of the small triangle. The desired ratio is $$\frac{3 \sqrt{6}}{6}= \frac{\sqrt{6}}{2}\Rightarrow (B)$$. 