2005 AMC 12B Problems/Problem 24

Problem

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 $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$ ?

$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}}$

Solution

Let the points be $(a,a^2)$ , $(b,b^2)$ and $(c,c^2)$ . Using elementary calculus concepts and the fact that they lie on $y = x^2$ ,

$a+b$ = the slope of $AB$ ,

$b+c$ = the slope of $BC$ ,

$a+c$ = the slope of $AC$ .

So the value that we need to find is simply the sum of the slopes of the three sides of the triangle divided by $2$ . WLOG, let $AB$ be the side that has the smallest angle with the positive $x$ -axis. Let $J$ be an arbitrary point with the coordinates $(1, 0)$ . Let us translate the triangle so $A$ is at the origin. Then $tan(BOJ) = 2$ . Using the fact that the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the value we now need to find is simply $\dfrac{tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)}{2}$ .

Using $tan(BOJ) = 2$ , and basic trigonometric identities, this simplifies to $\dfrac{3}{11}$ , so the answer is $3 + 11 = \boxed{\mathrm{(A)}\ 14}$

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2005 AMC 12B Problems/Problem 24

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