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2005 AMC 12B 第 24 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

24.

一个等边三角形的三个顶点都在抛物线 y=x2y = x^2 上，且其中一条边的斜率为 22。三个顶点的 xx-坐标之和为 mn\dfrac{m}{n}，其中 mm、nn 是互质正整数。m+nm + n 的值是多少？

All three vertices of an equilateral triangle are on the parabola y=x2,y = x^2, and one of its sides has a slope of 2.2. The xx-coordinates of the three vertices have a sum of mn,\dfrac{m}{n}, where mm and nn are relatively prime positive integers. What is the value of m+n?m + n?

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答案：A
知识点：抛物线三角恒等式等边三角形
难度评级：2300
小提示：

连接 (a,a2)(a, a^2) 和 (b,b2)(b, b^2) 的弦的斜率为 a+ba + b。

The chord joining (a,a2)(a, a^2) and (b,b2)(b, b^2) has slope a+ba + b

大提示：

三条边的斜率为 tan⁡θ\tan\theta 与 tan⁡(θ±60∘)\tan(\theta \pm 60^\circ)；它们的和是顶点横坐标之和的两倍。

The three side slopes are tan⁡θ\tan\theta and tan⁡(θ±60∘)\tan(\theta \pm 60^\circ); their sum is twice the vertex-sum

解答：

对顶点 (a,a2),(b,b2),(c,c2)(a, a^2), (b, b^2), (c, c^2)，一条边的斜率为 b2−a2b−a=a+b\dfrac{b^2 - a^2}{b - a} = a + b。三条边斜率相加得 (a+b)+(b+c)+(c+a)=2(a+b+c)=2⋅mn。 \begin{aligned} &(a+b) + (b+c) + (c+a) \\ &= 2(a + b + c) \\ &= 2 \cdot \dfrac{m}{n} \end{aligned}\text{。}

若一条边的方向角为 θ\theta，其斜率为 2=tan⁡θ2 = \tan\theta。等边三角形另外两条边的方向角为 θ±60∘\theta \pm 60^\circ，所以它们的斜率为 tan⁡(θ±60∘)=2±31∓23=−8±5311。 \begin{aligned} &\tan(\theta \pm 60^\circ) = \dfrac{2 \pm \sqrt3}{1 \mp 2\sqrt3} \\ &= -\dfrac{8 \pm 5\sqrt3}{11} \end{aligned}\text{。}

三个斜率和为 2−8+53112 - \dfrac{8 + 5\sqrt3}{11} −8−5311=- \dfrac{8 - 5\sqrt3}{11} = 22−1611=611\dfrac{22 - 16}{11} = \dfrac{6}{11}。

因此 a+b+c=12⋅611=311a + b + c = \dfrac12 \cdot \dfrac{6}{11} = \dfrac{3}{11}，所以 m+n=3+11=14m + n = 3 + 11 = 14。

所以正确答案是 A。

For vertices (a,a2),(b,b2),(c,c2),(a, a^2), (b, b^2), (c, c^2), the slope of a side is b2−a2b−a=a+b.\dfrac{b^2 - a^2}{b - a} = a + b. Adding the three side slopes, (a+b)+(b+c)+(c+a)=2(a+b+c)=2⋅mn. \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⁡θ.2 = \tan\theta. Because the triangle is equilateral, its sides make angles θ\theta and θ±60∘,\theta \pm 60^\circ, so the other two slopes are tan⁡(θ±60∘)=2±31∓23=−8±5311. \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 2−8+53112 - \dfrac{8 + 5\sqrt3}{11} −8−5311=- \dfrac{8 - 5\sqrt3}{11} = 22−1611=611.\dfrac{22 - 16}{11} = \dfrac{6}{11}.

Thus a+b+c=12⋅611=311,a + b + c = \dfrac12 \cdot \dfrac{6}{11} = \dfrac{3}{11}, so m+n=3+11=14.m + n = 3 + 11 = 14.

Thus, the correct answer is A.

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