Difference between revisions of "2013 AIME II Problems/Problem 4"
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− | ==Problem | + | ==Problem== |
In the Cartesian plane let <math>A = (1,0)</math> and <math>B = \left( 2, 2\sqrt{3} \right)</math>. Equilateral triangle <math>ABC</math> is constructed so that <math>C</math> lies in the first quadrant. Let <math>P=(x,y)</math> be the center of <math>\triangle ABC</math>. Then <math>x \cdot y</math> can be written as <math>\tfrac{p\sqrt{q}}{r}</math>, where <math>p</math> and <math>r</math> are relatively prime positive integers and <math>q</math> is an integer that is not divisible by the square of any prime. Find <math>p+q+r</math>. | In the Cartesian plane let <math>A = (1,0)</math> and <math>B = \left( 2, 2\sqrt{3} \right)</math>. Equilateral triangle <math>ABC</math> is constructed so that <math>C</math> lies in the first quadrant. Let <math>P=(x,y)</math> be the center of <math>\triangle ABC</math>. Then <math>x \cdot y</math> can be written as <math>\tfrac{p\sqrt{q}}{r}</math>, where <math>p</math> and <math>r</math> are relatively prime positive integers and <math>q</math> is an integer that is not divisible by the square of any prime. Find <math>p+q+r</math>. | ||
==Solution 1 == | ==Solution 1 == | ||
− | The distance from point <math>A</math> to point <math>B</math> is <math> \sqrt{13}</math>. The vector that starts at point A and ends at point B is given by <math>B - A = (1, 2\sqrt{3})</math>. Since the center of an equilateral triangle, <math>P</math>, is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equation for the perpendicular bisector to <math>\overline{AB}</math>. The line perpendicular to <math>\overline{AB}</math> through the midpoint, <math>M = (\dfrac{3}{2},\sqrt{3})</math>, <math>\overline{AB}</math> can be parameterized by <math> (\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}) t + (\dfrac{3}{2},\sqrt{3})</math>. At this point, it is useful to note that <math>\Delta BMP</math> is a 30-60-90 triangle with <math>\overline{MB}</math> measuring <math>\dfrac{\sqrt{13}}{2}</math>. This yields the length of <math>\overline{MP}</math> to be <math>\dfrac{\sqrt{13}}{2\sqrt{3}}</math>. Therefore, <math>P =( \dfrac{2\sqrt{3}}{\sqrt{13}},\dfrac{-1}{\sqrt{13}})(\dfrac{\sqrt{13}}{2\sqrt{3}}) + (\dfrac{3}{2},\sqrt{3}) = (\dfrac{5}{2}, \dfrac{5}{2\sqrt{3}})</math>. Therefore <math>xy = \dfrac{25\sqrt{3}}{12}</math> yielding an answer of <math> p + q + r = 25 + | + | The distance from point <math>A</math> to point <math>B</math> is <math> \sqrt{13}</math>. The vector that starts at point A and ends at point B is given by <math>B - A = (1, 2\sqrt{3})</math>. Since the center of an equilateral triangle, <math>P</math>, is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equation for the perpendicular bisector to <math>\overline{AB}</math>. The line perpendicular to <math>\overline{AB}</math> through the midpoint, <math>M = \left(\dfrac{3}{2},\sqrt{3}\right)</math>, <math>\overline{AB}</math> can be parameterized by <math>\left(\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}\right) t + \left(\dfrac{3}{2},\sqrt{3}\right)</math>. At this point, it is useful to note that <math>\Delta BMP</math> is a 30-60-90 triangle with <math>\overline{MB}</math> measuring <math>\dfrac{\sqrt{13}}{2}</math>. This yields the length of <math>\overline{MP}</math> to be <math>\dfrac{\sqrt{13}}{2\sqrt{3}}</math>. Therefore, <math>P =\left(\dfrac{2\sqrt{3}}{\sqrt{13}},\dfrac{-1}{\sqrt{13}}\right)\left(\dfrac{\sqrt{13}}{2\sqrt{3}}\right) + \left(\dfrac{3}{2},\sqrt{3}\right) = \left(\dfrac{5}{2}, \dfrac{5}{2\sqrt{3}}\right)</math>. Therefore <math>xy = \dfrac{25\sqrt{3}}{12}</math> yielding an answer of <math> p + q + r = 25 + 2 + 12 = \boxed{040}</math>. |
− | |||
==Solution 2== | ==Solution 2== | ||
Line 16: | Line 15: | ||
==Solution 3== | ==Solution 3== | ||
− | + | We can also consider the slopes of the lines. Midpoint <math>M</math> of <math>AB</math> has coordinates <math>\left(\frac{3}{2},\ \sqrt{3}\right)</math>. Because line <math>AB</math> has slope <math>2\sqrt{3}</math>, the slope of line <math>MP</math> is <math>-\frac{1}{2\sqrt{3}}</math> (Because of perpendicular slopes). | |
+ | |||
+ | Since <math>\Delta ABC</math> is equilateral, and since point <math>P</math> is the centroid, we can quickly calculate that <math>MP = \frac{\sqrt{39}}{6}</math>. Then, define <math>\Delta x</math> and <math>\Delta y</math> to be the differences between points <math>M</math> and <math>P</math>. Because of the slope, it is clear that <math>\Delta x = 2\sqrt{3} \Delta y</math>. | ||
+ | |||
+ | We can then use the Pythagorean Theorem on line segment <math>MP</math>: <math>MP^2 = \Delta x^2 + \Delta y^2</math> yields <math>\Delta y = -\frac{1}{2\sqrt{3}}</math> and <math>\Delta x = 1</math>, after substituting <math>\Delta x</math>. The coordinates of P are thus <math>\left(\frac{5}{2},\ \frac{5\sqrt{3}}{6}\right)</math>. Multiplying these together gives us <math>\frac{25\sqrt{3}}{12}</math>, giving us <math>\boxed{040}</math> as our answer. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Since <math>AC</math> will be segment <math>AB</math> rotated clockwise <math>60^{\circ}</math>, we can use a rotation matrix to find <math>C</math>. We first translate the triangle <math>1</math> unit to the left, so <math>A'</math> lies on the origin, and <math>B' = (1, 2\sqrt{3})</math>. Rotating clockwise <math>60^{\circ}</math> is the same as rotating <math>300^{\circ}</math> counter-clockwise, so our rotation matrix is <math>\begin{bmatrix} \cos{300^{\circ}} & -\sin{300^{\circ}}\\ \sin{300^{\circ}} & \cos{300^{\circ}}\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\\ \end{bmatrix}</math>. Then <math>C' = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\\ \end{bmatrix} \cdot \begin{bmatrix} 1\\ 2\sqrt{3}\\ \end{bmatrix} = \begin{bmatrix} \frac{7}{2}\\ \frac{\sqrt{3}}{2}\\ \end{bmatrix}</math>. Thus, <math>C = (\frac{9}{2}, \frac{\sqrt{3}}{2})</math>. Since the triangle is equilateral, the center of the triangle is the average of the coordinates of the vertices. Then <math>P = (\frac{1 + 2 + \frac{9}{2}}{3}, \frac{2\sqrt{3} + \frac{\sqrt{3}}{2}}{3}) = (\frac{5}{2}, \frac{5\sqrt{3}}{6})</math>. Our answer is <math>\frac{5}{2} \cdot \frac{5\sqrt{3}}{6} = \frac{25\sqrt{3}}{12}</math>. <math>25 + 3 + 12 = \boxed{40}</math> | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | We construct point <math>C</math> by drawing two circles with radius <math>r = AB = \sqrt{13}</math>. One circle will be centered at <math>A</math>, while the other is centered at <math>B</math>. The equations of the circles are: | ||
+ | |||
+ | <math>(x - 1)^2 + y^2 = 13</math> | ||
+ | |||
+ | <math>(x - 2)^2 + (y - 2\sqrt{3})^2 = 13</math> | ||
+ | |||
+ | Setting the LHS of each of these equations equal to each other and solving for <math>x</math> yields after simplification: | ||
+ | |||
+ | <math>x = \frac{15}{2} - 2\sqrt{3}y</math> | ||
+ | |||
+ | Plugging that into the first equation gives the following quadratic in <math>y</math> after simplification: | ||
+ | |||
+ | <math>y^2 - 2\sqrt{3}y + \frac{9}{4} = 0</math> | ||
+ | |||
+ | The quadratic formula gives <math>y = \frac{\sqrt{3}}{2}, \frac{3\sqrt{3}}{2}</math>. | ||
+ | |||
+ | Since <math>x > 0</math> and <math>x = \frac{15}{2} - 2\sqrt{3}y</math>, we pick <math>y = \frac{\sqrt{3}}{2}</math> in the hopes that it will give <math>x > 0</math>. Plugging <math>y</math> into the equation for <math>x</math> yields <math>x = \frac{9}{2}</math>. | ||
+ | |||
+ | Thus, <math>C(\frac{9}{2}, \frac{\sqrt{3}}{2})</math>. Averaging the coordinates of the vertices of equilateral triangle <math>ABC</math> will give the center of mass of the triangle. | ||
+ | |||
+ | Thus, <math>P(\frac{5}{2}, \frac{5\sqrt{3}}{6})</math>, and the product of the coordinates is <math>\frac{25\sqrt{3}}{12}</math>, so the desired quantity is <math>\boxed{040}</math>. | ||
+ | |||
+ | ==Solution 6== | ||
+ | Labeling our points and sketching a graph we get that <math>C</math> is to the right of <math>AB</math>. Of course, we need to find <math>C</math>. Note that the transformation from <math>A</math> to <math>B</math> is <math>[1,2\sqrt{3}]</math>, and if we imagine a height dropped to <math>AB</math> we see that a transformation from the midpoint <math>(\frac{3}{2},\sqrt {3})</math> to <math>C</math> is basically the first transformation, with <math>\frac{\sqrt{3}}{2}</math> the magnitude and the x and y switched– then multiply the new y by -1. Then, applying this transformation of <math>[3,\frac{-\sqrt{3}}{2}]</math> we get that <math>C=(\frac{9}{2},\frac{\sqrt{3}}{2})</math> which means that <math>P=(\frac{5}{2},\frac{5\sqrt{3}}{6})</math>. Then our answer is <math>\boxed{40}</math>. | ||
− | + | ==Solution 7== | |
+ | Transform this into the complex plane and let <math>a=1, b=2+2\sqrt3 i</math>. We know that 3 complex numbers <math>a,b,c</math> form an equilateral triangle if <math>a^2+b^2+c^2=ab+bc+ac</math>, so plugging in our values of <math>a,b</math>, we get <math>8\sqrt3 i - 7 +c^2 = 2+2\sqrt3 i + (3+2\sqrt 3i)c.</math> Solving for <math>c</math> using Wolfram Alpha, we find that the solutions are <math>c=\frac 92 + \frac{i\sqrt3}{2}, -\frac 32 + \frac{3i\sqrt3}{2}</math>. The first one is in the first quadrant, so <math>C\left( \frac 92, \frac{\sqrt3}{2} \right)</math>. The center is the average of the coordinates and we find that it is <math>\left(\frac{5}2, \frac{5\sqrt3}{6} \right)</math>. Then <math>xy = \frac{25\sqrt3}{12} \implies 40</math>. | ||
− | + | -bobthegod78, krwang, and Simplest14 | |
==See Also== | ==See Also== | ||
{{AIME box|year=2013|n=II|num-b=3|num-a=5}} | {{AIME box|year=2013|n=II|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:10, 31 January 2024
Contents
Problem
In the Cartesian plane let and
. Equilateral triangle
is constructed so that
lies in the first quadrant. Let
be the center of
. Then
can be written as
, where
and
are relatively prime positive integers and
is an integer that is not divisible by the square of any prime. Find
.
Solution 1
The distance from point to point
is
. The vector that starts at point A and ends at point B is given by
. Since the center of an equilateral triangle,
, is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equation for the perpendicular bisector to
. The line perpendicular to
through the midpoint,
,
can be parameterized by
. At this point, it is useful to note that
is a 30-60-90 triangle with
measuring
. This yields the length of
to be
. Therefore,
. Therefore
yielding an answer of
.
Solution 2
Rather than considering the Cartesian plane, we use complex numbers. Thus A is 1 and B is .
Recall that a rotation of radians counterclockwise is equivalent to multiplying a complex number by
, but here we require a clockwise rotation, so we multiply by
to obtain C. Upon averaging the coordinates of A, B, and C, we obtain the coordinates of P, viz.
.
Therefore is
and the answer is
.
Solution 3
We can also consider the slopes of the lines. Midpoint of
has coordinates
. Because line
has slope
, the slope of line
is
(Because of perpendicular slopes).
Since is equilateral, and since point
is the centroid, we can quickly calculate that
. Then, define
and
to be the differences between points
and
. Because of the slope, it is clear that
.
We can then use the Pythagorean Theorem on line segment :
yields
and
, after substituting
. The coordinates of P are thus
. Multiplying these together gives us
, giving us
as our answer.
Solution 4
Since will be segment
rotated clockwise
, we can use a rotation matrix to find
. We first translate the triangle
unit to the left, so
lies on the origin, and
. Rotating clockwise
is the same as rotating
counter-clockwise, so our rotation matrix is
. Then
. Thus,
. Since the triangle is equilateral, the center of the triangle is the average of the coordinates of the vertices. Then
. Our answer is
.
Solution 5
We construct point by drawing two circles with radius
. One circle will be centered at
, while the other is centered at
. The equations of the circles are:
Setting the LHS of each of these equations equal to each other and solving for yields after simplification:
Plugging that into the first equation gives the following quadratic in after simplification:
The quadratic formula gives .
Since and
, we pick
in the hopes that it will give
. Plugging
into the equation for
yields
.
Thus, . Averaging the coordinates of the vertices of equilateral triangle
will give the center of mass of the triangle.
Thus, , and the product of the coordinates is
, so the desired quantity is
.
Solution 6
Labeling our points and sketching a graph we get that is to the right of
. Of course, we need to find
. Note that the transformation from
to
is
, and if we imagine a height dropped to
we see that a transformation from the midpoint
to
is basically the first transformation, with
the magnitude and the x and y switched– then multiply the new y by -1. Then, applying this transformation of
we get that
which means that
. Then our answer is
.
Solution 7
Transform this into the complex plane and let . We know that 3 complex numbers
form an equilateral triangle if
, so plugging in our values of
, we get
Solving for
using Wolfram Alpha, we find that the solutions are
. The first one is in the first quadrant, so
. The center is the average of the coordinates and we find that it is
. Then
.
-bobthegod78, krwang, and Simplest14
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.