Difference between revisions of "2013 AIME II Problems/Problem 4"
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==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 + 3 + 12 = \boxed{040}</math>. | + | 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 + 3 + 12 = \boxed{040}</math>. |
− | |||
==Solution 2== | ==Solution 2== |
Revision as of 19:45, 8 December 2017
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
.
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
.
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.