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

(Solution)
(Solution)
Line 5: Line 5:
 
<math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math>
 
<math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math>
 
== Solution ==
 
== Solution ==
 
Let the points be <math>(a,a^2)</math>, <math>(b,b^2)</math> and <math>(c,c^2)</math>.
 
  
 
<center><asy>
 
<center><asy>
Line 20: Line 18:
 
B = (b,f(b));
 
B = (b,f(b));
 
C = (c,f(c));
 
C = (c,f(c));
 +
J = (1,0);
 
draw(graph(f,-2,2));
 
draw(graph(f,-2,2));
 
draw((-2,0)--(2,0),Arrows);
 
draw((-2,0)--(2,0),Arrows);
Line 29: Line 28:
 
dot("$B(b,b^2)$", B, E);
 
dot("$B(b,b^2)$", B, E);
 
dot("$C(c,c^2)$", C, W);
 
dot("$C(c,c^2)$", C, W);
 +
dot("$J(1,0)$", 1,0);
 
</asy></center>
 
</asy></center>
 
Using the slope formula and differences of squares, we find:  
 
Using the slope formula and differences of squares, we find:  

Revision as of 18:20, 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

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));
J = (1,0);
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);
dot("$J(1,0)$", 1,0);
 (Error compiling LaTeX. J = (1,0);
^
73991b8a40fc6ed513cd837b5c80627f57ffdee2.asy: 14.1: no matching variable 'J')

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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