Difference between revisions of "2013 AIME II Problems/Problem 4"
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==Solution 4== | ==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> | 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>. | ||
==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 01:47, 15 March 2016
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 .
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.