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Difference between revisions of "2005 AMC 12B Problems/Problem 24"

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Let the points be <math>(a,a^2)</math>, <math>(b,b^2)</math> and <math>(c,c^2)</math>.  
 
Let the points be <math>(a,a^2)</math>, <math>(b,b^2)</math> and <math>(c,c^2)</math>.  
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[asy]
 
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import graph;
 
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real f(real x) {return x^2;}
 
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unitsize(1 cm);
 
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pair A, B, C;
 
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real a, b, c;
 
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a = (-5*sqrt(3) + 11)/11;
 
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b = (5*sqrt(3) + 11)/11;
 
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c = -19/11;
 
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A = (a,f(a));
 
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B = (b,f(b));
 
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C = (c,f(c));
 
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draw(graph(f,-2,2));
 
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draw((-2,0)--(2,0),Arrows);
 
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draw((0,-0.5)--(0,4),Arrows);
 
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draw(A--B--C--cycle);
 
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label("<math>x</math>", (2,0), NE);
 
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label("<math>y</math>", (0,4), NE);
 
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dot("<math>A(a,a^2)</math>", A, S);
 
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dot("<math>B(b,b^2)</math>", B, E);
 
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dot("<math>C(c,c^2)</math>", C, W);[/asy]
 
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Revision as of 18:09, 30 August 2013

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 the slope formula and differences of squares, we find:

$a+b$ = 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 the sum of the slopes of the three sides of the triangle divided by $2$. Without loss of generality, let $AB$ be the side that has the smallest angle with the positive $x$-axis. Let $J$ be an arbitrary point with the coordinates $(1, 0)$. Translate the triangle so $A$ is at the origin. Then $tan(BOJ) = 2$. 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 $\dfrac{tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)}{2}$.

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

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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