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

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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>
 
<math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math>
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== Solution ==
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== Solution 1 (Complex numbers)==
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Let the three points be at <math>A = (x_1, x_1^2)</math>, <math>B = (x_2, x_2^2)</math>, and <math>C = (x_3, x_3^2)</math>, such that the slope between the first two is <math>2</math>.
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Then, we have <math>\textrm{Slope of }AC = \frac{x_1 - x_3}{x_1^2 - x_3^2} = x_1 + x_3</math>. Similarly, the slope of <math>BC</math> is <math>x_2 + x_3</math>, and the slope of <math>AB</math> is <math>x_1 + x_2 = 2</math>. The desired sum is <math>x_1 + x_2 + x_3</math>, which is equal to the sum of the slopes divided by <math>2</math>.
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To find the slope of <math>AC</math>, we note that it comes at a <math>60^{\circ}</math> angle with <math>AB</math>. Thus, we can find the slope of <math>AC</math> by multiplying the two complex numbers <math>1 + 2i</math> and <math>1 + \sqrt{3}i</math>. What this does is generate the complex number that is at a <math>60^{\circ}</math> angle with the complex number <math>1 + 2i</math>. Then, we can find the slope of the line between this new complex number and the origin:
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<cmath>(1+2i)(1+\sqrt{3}i)</cmath>
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<cmath> = 1 - 2\sqrt{3} + 2i + \sqrt{3}i</cmath>
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<cmath>\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}</cmath>
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<cmath> = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}</cmath>
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<cmath> = \frac{8 + 5\sqrt{3}}{-11}</cmath>
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<cmath> = \frac{-8 - 5\sqrt{3}}{11}.</cmath>
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== Solution 2 ==
  
 
<center><asy>
 
<center><asy>

Revision as of 02:10, 26 November 2022

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 1 (Complex numbers)

Let the three points be at $A = (x_1, x_1^2)$, $B = (x_2, x_2^2)$, and $C = (x_3, x_3^2)$, such that the slope between the first two is $2$.

Then, we have $\textrm{Slope of }AC = \frac{x_1 - x_3}{x_1^2 - x_3^2} = x_1 + x_3$. Similarly, the slope of $BC$ is $x_2 + x_3$, and the slope of $AB$ is $x_1 + x_2 = 2$. The desired sum is $x_1 + x_2 + x_3$, which is equal to the sum of the slopes divided by $2$.

To find the slope of $AC$, we note that it comes at a $60^{\circ}$ angle with $AB$. Thus, we can find the slope of $AC$ by multiplying the two complex numbers $1 + 2i$ and $1 + \sqrt{3}i$. What this does is generate the complex number that is at a $60^{\circ}$ angle with the complex number $1 + 2i$. Then, we can find the slope of the line between this new complex number and the origin: \[(1+2i)(1+\sqrt{3}i)\] \[= 1 - 2\sqrt{3} + 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}\] \[= \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}\] \[= \frac{8 + 5\sqrt{3}}{-11}\] \[= \frac{-8 - 5\sqrt{3}}{11}.\]

Solution 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 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 (Problems • Answer Key • Resources)
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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