Question

一个等边三角形的三个顶点都在抛物线 $y = x^2$ 上，且其中一条边的斜率为 $2$ 。三个顶点的 $x$ -坐标之和为 $\dfrac{m}{n}$ ，其中 $m$ 、 $n$ 是互质正整数。求 $m + n$ ？

All three vertices of an equilateral triangle are on the parabola $y = x^2,$ and one of its sides has a slope of $2.$ The $x$ -coordinates of the three vertices have a sum of $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n?$

$14$

$15$

$16$

$17$

$18$

答案：A

知识点：抛物线三角恒等式等边三角形

难度评级：2300

解答：

对顶点 $(a, a^2), (b, b^2), (c, c^2)$ ，一条边的斜率为 $\dfrac{b^2 - a^2}{b - a} = a + b$ 。三条边斜率相加得 $\begin{aligned} &(a+b) + (b+c) + (c+a) \\ &= 2(a + b + c) \\ &= 2 \cdot \dfrac{m}{n}. \end{aligned}$

若一条边的方向角为 $\theta$ ，其斜率为 $2 = \tan\theta$ 。等边三角形另外两条边的方向角为 $\theta \pm 60^\circ$ ，所以它们的斜率为 $\begin{aligned} &\tan(\theta \pm 60^\circ) = \dfrac{2 \pm \sqrt3}{1 \mp 2\sqrt3} \\ &= -\dfrac{8 \pm 5\sqrt3}{11}. \end{aligned}$

三个斜率和为 $2 - \dfrac{8 + 5\sqrt3}{11}$ $- \dfrac{8 - 5\sqrt3}{11} =$ $\dfrac{22 - 16}{11} = \dfrac{6}{11}$ 。

因此 $a + b + c = \dfrac12 \cdot \dfrac{6}{11} = \dfrac{3}{11}$ ，所以 $m + n = 3 + 11 = 14$ 。

所以正确答案是 A。

For vertices $(a, a^2), (b, b^2), (c, c^2),$ the slope of a side is $\dfrac{b^2 - a^2}{b - a} = a + b.$ Adding the three side slopes, $\begin{aligned} &(a+b) + (b+c) + (c+a) \\ &= 2(a + b + c) \\ &= 2 \cdot \dfrac{m}{n}. \end{aligned}$

One side has slope $2 = \tan\theta.$ Because the triangle is equilateral, its sides make angles $\theta$ and $\theta \pm 60^\circ,$ so the other two slopes are $\begin{aligned} &\tan(\theta \pm 60^\circ) = \dfrac{2 \pm \sqrt3}{1 \mp 2\sqrt3} \\ &= -\dfrac{8 \pm 5\sqrt3}{11}. \end{aligned}$

The sum of the three slopes is $2 - \dfrac{8 + 5\sqrt3}{11}$ $- \dfrac{8 - 5\sqrt3}{11} =$ $\dfrac{22 - 16}{11} = \dfrac{6}{11}.$

Thus $a + b + c = \dfrac12 \cdot \dfrac{6}{11} = \dfrac{3}{11},$ so $m + n = 3 + 11 = 14.$

Thus, the correct answer is A.

2005 AMC 12B 第 24 题

答案：A

难度评级：2300

解答：

其他年份的第 24 题