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Difference between revisions of "2005 AMC 12B Problems/Problem 24"
m (slope formula is wrong for solution 1)
 
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{{empty}}
 
 
== Problem ==
 
== Problem ==
  
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All three vertices of an equilateral triangle are on the parabola , and one of its sides has a slope of . The -coordinates of the three vertices have a sum of , where  and  are relatively prime positive integers. What is the value of ?
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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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{{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? }}
 
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== Solution ==
 
  
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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,
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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>
  
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a+b = the slope of AB,
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== Solution 1 (Complex numbers)==
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b+c = the slope of BC,
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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>, and <math>A</math> is the point with the least <math>y</math>-coordinate.
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a+c = the slope of AC.
 
  
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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,
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Then, we have <math>\textrm{Slope of }AC = \frac{x_1^2 - x_3^2}{x_1 - x_3} = 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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                                        (tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)) / 2.
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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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The slope <math>BC</math> can also be solved similarly, noting that it makes a <math>120^{\circ}</math> angle with <math>AB</math>:
 +
<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{\sqrt{3} - 2}{-2\sqrt{3} - 1}</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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Using tan(BOJ) = 2, and basic trigonometric identities, this simplifies to 3/11, so the answer is 3 + 11 = 14 (A).  
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At this point, we start to notice a pattern: This expression is equal to <math>\frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}</math>, except the numerators of the first fractions are conjugates! Notice that this means that when we multiply out, the rational term will stay the same, but the coefficient of <math>\sqrt{3}</math> will have its sign switched. This means that the two complex numbers are conjugates, so their irrational terms cancel out.
  
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Our sum is simply <math>2 - 2\cdot\frac{8}{11} = \frac{6}{11}</math>, and thus we can divide by <math>2</math> to obtain <math>\frac{3}{11}</math>, which gives the answer <math>\boxed{\mathrm{(A)}\ 14}</math>.
  
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~mathboy100
  
 +
== Solution 2 ==
 +
 +
<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>
 +
Using the slope formula and differences of squares, we find:
 +
 +
<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>a+c</math> = the slope of <math>AC</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 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}}
 +
{{MAA Notice}}

Latest revision as of 08:10, 1 August 2023

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 $2$, and $A$ is the point with the least $y$-coordinate.

Then, we have $\textrm{Slope of }AC = \frac{x_1^2 - x_3^2}{x_1 - x_3} = x_1 + x_3$. 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: \[(1+2i)(1+\sqrt{3}i)\] \[= 1 - 2\sqrt{3} + 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}\] \[= \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}\] \[= \frac{8 + 5\sqrt{3}}{-11}\] \[= \frac{-8 - 5\sqrt{3}}{11}.\] The slope $BC$ can also be solved similarly, noting that it makes a $120^{\circ}$ angle with $AB$: \[(1+2i)(-1+\sqrt{3}i)\] \[= -1 - 2\sqrt{3} - 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{\sqrt{3} - 2}{-2\sqrt{3} - 1}\] \[= \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \cdot \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}\]

At this point, we start to notice a pattern: This expression is equal to $\frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}$, except the numerators of the first fractions are conjugates! Notice that this means that when we multiply out, the rational term will stay the same, but the coefficient of $\sqrt{3}$ will have its sign switched. This means that the two complex numbers are conjugates, so their irrational terms cancel out.

Our sum is simply $2 - 2\cdot\frac{8}{11} = \frac{6}{11}$, and thus we can divide by $2$ to obtain $\frac{3}{11}$, which gives the answer $\boxed{\mathrm{(A)}\ 14}$.

~mathboy100

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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