Question

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 $m + n?$

$14$

$15$

$16$

$17$

$18$

Solution:

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, $(a+b) + (b+c) + (c+a) = 2(a + b + c) = 2 \cdot \dfrac{m}{n}.$

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 $\tan(\theta \pm 60^\circ) = \dfrac{2 \pm \sqrt3}{1 \mp 2\sqrt3} = -\dfrac{8 \pm 5\sqrt3}{11}.$

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 Problem 24

Solution:

Problem 24 in Other Years