2005 AMC 12B 第 24 题

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。

m + n

的值是多少？

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

小提示：

连接

(a, a^2)

和

(b, b^2)

的弦的斜率为

a + b

。

The chord joining

(a, a^2)

and

(b, b^2)

has slope

a + b

大提示：

三条边的斜率为

\tan\theta

与

\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\theta

and

\tan(\theta \pm 60^\circ)

; their sum is twice the vertex-sum

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}\text{。}

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}\text{。}

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

小提示：

大提示：

解答：

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