Difference between revisions of "2013 AIME II Problems/Problem 4"
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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 iff <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>. | 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 iff <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 and | + | -bobthegod78, krwang, and Simplest14 |
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==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}} |
Revision as of 22:46, 7 November 2022
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 iff
, 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.