2005 AMC 12B Problems/Problem 24
Problem
All three vertices of an equilateral triangle are on the parabola 
, and one of its sides has a slope of 
. The 
-coordinates of the three vertices have a sum of 
, where 
and 
are relatively prime positive integers. What is the value of 
?
Solution
![[asy] import graph; real f(real x) {return x^2;} unitsize(1 cm); pair A, B, C; real a, b, c; a = (-5*sqrt(3) + 11)/11; b = (5*sqrt(3) + 11)/11; c = -19/11; A = (a,f(a)); B = (b,f(b)); C = (c,f(c)); draw(graph(f,-2,2)); draw((-2,0)--(2,0),Arrows); draw((0,-0.5)--(0,4),Arrows); draw(A--B--C--cycle); label("$x$", (2,0), NE); label("$y$", (0,4), NE); dot("$A(a,a^2)$", A, S); dot("$B(b,b^2)$", B, E); dot("$C(c,c^2)$", C, W); [/asy]](http://latex.artofproblemsolving.com/1/6/1/16184f744e08f706cf208e04e55092dcd13671ad.png)
Using the slope formula and differences of squares, we find:
= the slope of 
,
= the slope of 
,
= the slope of 
.
So the value that we need to find is the sum of the slopes of the three sides of the triangle divided by . Without loss of generality, let be the side that has the smallest angle with the positive -axis. Let 
be an arbitrary point with the coordinates 
. Translate the triangle so 
is at the origin. Then 
. Since the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the answer is 
.
Using , and the tangent addition formula, this simplifies to 
, so the answer is 
See also
| 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 | |
| All AMC 12 Problems and Solutions | |
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