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>. |
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==Solution 2== | ==Solution 2== | ||
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==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>. | + | 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>. | 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>. | ||
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==Solution 6== | ==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 | + | 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>. |
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+ | ==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>. | ||
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+ | -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 19:06, 17 October 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 |
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