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

Revision as of 20:30, 29 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 $(c,c^2)$ .

[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]

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 $3 + 11 = \boxed{\mathrm{(A)}\ 14}$

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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Difference between revisions of "2005 AMC 12B Problems/Problem 24"

Revision as of 20:30, 29 August 2013

Problem

Solution

See also

@@ Line 6: / Line 6: @@
 == Solution ==
-Let the points be <math>(a,a^2)</math>, <math>(b,b^2)</math> and <math>(c,c^2)</math>. Using the slope formula and differences of squares, we find:
+Let the points be <math>(a,a^2)</math>, <math>(b,b^2)</math> and <math>(c,c^2)</math>.
+[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("<math>x</math>", (2,0), NE);
+label("<math>y</math>", (0,4), NE);
+dot("<math>A(a,a^2)</math>", A, S);
+dot("<math>B(b,b^2)</math>", B, E);
+dot("<math>C(c,c^2)</math>", C, W);[/asy]
+Using the slope formula and differences of squares, we find:
 <math>a+b</math> = the slope of <math>AB</math>,