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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>
−
== Solution ==
 
  
−
Let the points be <math>(a,a^2)</math>, <math>(b,b^2)</math> and <math>(c,c^2)</math>.  
+
== Solution 1 (Complex numbers)==
 +
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.
  
 +
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>.
  
 +
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:
 +
<cmath>(1+2i)(1+\sqrt{3}i)</cmath>
 +
<cmath> = 1 - 2\sqrt{3} + 2i + \sqrt{3}i</cmath>
 +
<cmath>\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}</cmath>
 +
<cmath> = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}</cmath>
 +
<cmath> = \frac{8 + 5\sqrt{3}}{-11}</cmath>
 +
<cmath> = \frac{-8 - 5\sqrt{3}}{11}.</cmath>
 +
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>
 +
<cmath> = -1 - 2\sqrt{3} - 2i + \sqrt{3}i</cmath>
 +
<cmath>\textrm{Slope } = \frac{\sqrt{3} - 2}{-2\sqrt{3} - 1}</cmath>
 +
<cmath> = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \cdot \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}</cmath>
 +
 +
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.
 +
 +
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>.
 +
 +
~mathboy100
 +
 +
== Solution 2 ==
 +
 +
<center><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></center>
 
Using the slope formula and differences of squares, we find:  
 
Using the slope formula and differences of squares, we find:  
  
Line 17: Line 61:
 
<math>a+c</math> = the slope of <math>AC</math>.  
 
<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>.  
+
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 trig 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 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 ==
 
{{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}}

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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