Difference between revisions of "2013 AIME II Problems/Problem 4"
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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>. | |
− | + | 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>040</math> as our answer. | |
==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 11:26, 2 March 2014
Problem 4
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.
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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