Question

一条经过原点的直线同时与直线 $x = 1$ 和直线相交。这三条直线围成一个等边三角形。这个三角形的周长是多少？ $y=1+ \dfrac{\sqrt{3}}{3} x.$

A line that passes through the origin intersects both the line $x = 1$ and the line $y=1+ \dfrac{\sqrt{3}}{3} x.$ The three lines create an equilateral triangle. What is the perimeter of the triangle?

$2\sqrt{6}$

$2 + 2\sqrt{3}$

$6$

$3 + 2\sqrt{3}$

$6 + \dfrac{\sqrt{3}}{3}$

答案：D

知识点：坐标几何等边三角形斜率

难度评级：1540

解答：

因为等边三角形的一边是竖直线，所以其对称轴为水平线。

经过原点的第三条边斜率应为 $-\dfrac{\sqrt{3}}{3}$ 。

在这条竖直线上，另外两条斜边的纵坐标之差就是边长。

令 $x = 1$ 。

两个 $y$ 值分别为 $1 + \dfrac{\sqrt{3}}{3}$ 和 $-\dfrac{\sqrt{3}}{3}$ 。它们的差为 $1 + \dfrac{2\sqrt{3}}{3}$ ，所以周长为因此正确答案是 D。 $3 \cdot \left(1 + \dfrac{2\sqrt{3}}{3}\right) = 3 + 2\sqrt{3}.$

Since one of the sides of the equilateral triangle is a vertical line, the line of symmetry perpendicular to this side must be horizontal.

This means that the slope of the third side must be opposite the slope of the second side, which would be $-\dfrac{\sqrt{3}}{3}.$

To find the perimeter, we only need to find the length of one of the sides of the triangle.

We can plug in $x = 1$ into the two other equations to get the two vertices on the vertical line.

The two $y$ -values are $1 + \dfrac{\sqrt{3}}{3}$ and $-\dfrac{\sqrt{3}}{3}.$ Their difference is $1 + \dfrac{2\sqrt{3}}{3},$ which makes the perimeter $3 \cdot \left(1 + \dfrac{2\sqrt{3}}{3}\right) = 3 + 2\sqrt{3}.$ Thus, D is the correct answer.

2015 AMC 10A 第 17 题

答案：D

难度评级：1540

解答：

其他年份的第 17 题