# 2005 AMC 12B Problems/Problem 24

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## Problem

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## 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,

(tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)) / 2.

Using tan(BOJ) = 2, and basic trigonometric identities, this simplifies to 3/11, so the answer is 3 + 11 = 14 (A).