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

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== Solution ==
 
== Solution ==
  
Let the points be <math>(a,a^2)</math>, <math>(b,b^2)</math> and <math>(c,c^2)</math>. Using elementary calculus concepts and the fact that they lie on <math>y = x^2</math>,
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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:
  
<math>a+b</math> = the slope of <math>AB</math>,
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<math>a+b</math> = the slope of <math>AB</math>,  
  
<math>b+c</math> = the slope of <math>BC</math>,
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<math>b+c</math> = the slope of <math>BC</math>,  
  
<math>a+c</math> = the slope of <math>AC</math>.
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<math>a+c</math> = the slope of <math>AC</math>.  
  
So the value that we need to find is simply the sum of the slopes of the three sides of the triangle divided by <math>2</math>. WLOG, 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>. Let us translate the triangle so <math>A</math> is at the origin. Then <math>tan(BOJ) = 2</math>. Using the fact that the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the value we now need to find is simply <math>\dfrac{tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)}{2}</math>.  
+
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>.  
  
Using <math>tan(BOJ) = 2</math>, and basic trigonometric identities, this simplifies to <math>\dfrac{3}{11}</math>, so the answer is <math>3 + 11 = \boxed{\mathrm{(A)}\ 14}</math>
+
Using <math>tan(BOJ) = 2</math>, and basic trig identities, this simplifies to <math>\dfrac{3}{11}</math>, so the answer is <math>3 + 11 = \boxed{\mathrm{(A)}\ 14}</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2005|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2005|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:26, 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 $(c,c^2)$. 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 (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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