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

m (Fixed LaTeX)
Line 1: Line 1:
{{empty}}
 
 
== Problem ==
 
== 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?  
+
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>?  
 +
 
 +
<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 (a,a^2), (b,b^2) and (c,c^2). Using elementary calculus concepts and the fact that they lie on y = x^2,
+
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>,
 
 
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 simply the sum of the slopes of the three sides of the triangle divided by 2. WLOG, 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). Let us translate the triangle so A is at the origin. Then tan(BOJ) = 2. 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,
 
  
                                        (tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)) / 2.
+
<math>a+b</math> = the slope of <math>AB</math>,
  
Using tan(BOJ) = 2, and basic trigonometric identities, this simplifies to 3/11, so the answer is 3 + 11 = 14 (A).
+
<math>b+c</math> = the slope of <math>BC</math>,
  
 +
<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>.
  
 +
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>
  
 
== See also ==
 
== See also ==
* [[2005 AMC 12B Problems]]
+
{{AMC12 box|year=2005|ab=B|num-b=23|num-a=25}}

Revision as of 13:52, 4 July 2011

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 elementary calculus concepts and the fact that they lie on $y = x^2$,

$a+b$ = 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 simply the sum of the slopes of the three sides of the triangle divided by $2$. WLOG, let $AB$ be the side that has the smallest angle with the positive $x$-axis. Let $J$ be an arbitrary point with the coordinates $(1, 0)$. Let us translate the triangle so $A$ is at the origin. Then $tan(BOJ) = 2$. 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 $\dfrac{tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)}{2}$.

Using $tan(BOJ) = 2$, and basic trigonometric 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
Invalid username
Login to AoPS