2005 AMC 12B Problem 24

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

The chord joining

(a, a^2)

and

(b, b^2)

has slope

a + b

The three side slopes are

\tan\theta

and

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

; their sum is twice the vertex-sum

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.

Problem 24 in other years