AMC12 2010 A

AMC12 2010 A · Q17

AMC12 2010 A · Q17. It mainly tests Triangles (properties), Area & perimeter.

Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?

等角六边形 $ABCDEF$ 有边长 $AB=CD=EF=1$ 且 $BC=DE=FA=r$。$\triangle ACE$ 的面积是该六边形面积的 $70\%$。所有可能的 $r$ 值之和是多少？

(A) \frac{4\sqrt{3}}{3} \frac{4\sqrt{3}}{3}

(B) \frac{10}{3} \frac{10}{3}

(D) \frac{17}{4} \frac{17}{4}

(E) 6 6

Answer

Correct choice: (E)

正确答案：（E）

Solution

It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\triangle ABC$, we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$. Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle. If we extend $BC$, $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$, $BC$ and $DE$ meet at $Y$, and $DE$ and $FA$ meet at $Z$, we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$, $CDY$ and $EFZ$, of side length $1$. The area of $ABCDEF$ is therefore $\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$. Based on the initial conditions, \[\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)\] Simplifying this gives us $r^2-6r+1 = 0$. By Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{\textbf{(E)}\ 6}$.

显然 $\triangle ACE$ 是等边三角形。对 $\triangle ABC$ 使用余弦定理，得到 $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$。因此由等边三角形面积公式，$\triangle ACE$ 的面积为 $\frac{\sqrt{3}}{4}(r^2+r+1)$。延长 $BC$、$DE$ 和 $FA$，使得 $FA$ 与 $BC$ 交于 $X$，$BC$ 与 $DE$ 交于 $Y$，$DE$ 与 $FA$ 交于 $Z$。可以发现六边形 $ABCDEF$ 可由边长为 $r+2$ 的等边三角形 $XYZ$ 去掉三个边长为 $1$ 的等边三角形 $ABX$、$CDY$ 和 $EFZ$ 得到。因此 $ABCDEF$ 的面积为 $\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$。由题设条件， \[\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)\] 化简得 $r^2-6r+1 = 0$。由韦达定理可知所有可能的 $r$ 值之和为 $\boxed{\textbf{(E)}\ 6}$。

Topics