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

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== Problem ==
 
== Problem ==
{{problem}}
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All three vertices of an equilateral triangle are on the parabola <math>y = x^2</math>, and one of its sides has a slope of <math>2</math>. The <math>x</math>-coordinates of the three vertices have a sum of <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is the value of <math>m + n</math>?
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<math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math>
 
== Solution ==
 
== Solution ==
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<center><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("$x$", (2,0), NE);
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label("$y$", (0,4), NE);
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dot("$A(a,a^2)$", A, S);
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dot("$B(b,b^2)$", B, E);
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dot("$C(c,c^2)$", C, W);
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</asy></center>
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Using the slope formula and differences of squares, we find:
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<math>a+b</math> = the slope of <math>AB</math>,
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<math>b+c</math> = the slope of <math>BC</math>,
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<math>a+c</math> = the slope of <math>AC</math>.
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So the value that we need to find is the sum of the slopes of the three sides of the triangle divided by <math>2</math>. Without loss of generality, let <math>AB</math> be the side that has the smallest angle with the positive <math>x</math>-axis. Let <math>J</math> be an arbitrary point with the coordinates <math>(1, 0)</math>. Translate the triangle so <math>A</math> is at the origin. Then <math>\tan(BOJ) = 2</math>. 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 <math>\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}</math>.
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Using <math>\tan(BOJ) = 2</math>, and the tangent addition formula, this simplifies to <math>\dfrac{3}{11}</math>, so the answer is <math>3 + 11 = \boxed{\mathrm{(A)}\ 14}</math>
  
 
== See also ==
 
== See also ==
* [[2005 AMC 12B Problems]]
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{{AMC12 box|year=2005|ab=B|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 14:12, 15 July 2018

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

[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 the tangent addition formula, this simplifies to $\dfrac{3}{11}$, so the answer is $3 + 11 = \boxed{\mathrm{(A)}\ 14}$

See also

2005 AMC 12B (ProblemsAnswer KeyResources)
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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