# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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Simpler than it Sounds. Topic: Complex Numbers. Level: AMC/AIME. November 24th, 2005

Problem #1: Find the coordinates of the point rotated around the point by radians counterclockwise.

Solution: Consider these points in the complex plane (-axis becomes real part, -axis becomes imaginary part).

We want to rotate around . Rewrite these in form, where is the magnitude and is the angle. We also have by the Euler Formula.

So, to simplify things, we shift to the origin, and now we want to rotate . We notice that rotation by simply adds to the angle, resulting in .

Shifting back to the “real” origin, our rotated point becomes in the complex plane and in the Cartesian plane. QED.

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Problem #2: Given that and are two vertices of an equilateral triangle, find all possible values for the third point, .

Solution: Convert them to complex numbers – and .

Notice that if we rotate around by or radians, we get the only two possible points for ( as a complex number).

So applying the same rotation method as in the first problem, we have or . We can translate this back to the Cartesian plane, but it’s just a mess of algebra, not essential to understanding the method of rotation in the complex plane. QED.

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Practice Problem #1: Find the value of rotated by radians counterclockwise.

Practice Problem #2: (WOOT Class) Show that, given three complex numbers that lie on the unit circle, the orthocenter of the triangle formed by them is (hint: If two complex numbers are perpendicular, is purely imaginary… prove it).

Practice Problem #3: (WOOT Message Board) Let be the center of a circle . Points on are chosen such that the triangles are equilateral. Let be the midpoints of , respectively. Prove that triangle is equilateral.

### 4 Responses to “Simpler than it Sounds. Topic: Complex Numbers. Level: AMC/AIME.”

1. chess64 Says:

1. A=3+i=(\sqrt{10}, \arcsin {1/3}). Rotating by \pi/2 is just adding \pi/2 to \theta = \arcsin (1/3). Now we need to convert back to rectangular form.

\cos (\theta) = \cos (\arcsin (\theta))\cos (\pi/2) – \sin (\arcsin (\theta))\sin \pi/2
\cos (\theta) = -1/3
\sin (\theta) = \sqrt{1-1/9}=\sqrt{8}/3

So the new point is (-1/3, \sqrt{8}{3}). I hope I got it right…

Not quite, the angle isn’t arcsin(1/3), it’s actually arctan(1/3). Hint: Rotation by pi/2 = Multiplying by e^(i*pi/2) = Multiplying by i.

3. chess64 Says:

Oops…so it’s just multiplying by i? Then A’ = -1 + 3i = (-1,3)?

4. Mathematical Food for Thought Says:

[...] Practice Problem #2: In an acute angled triangle , . is the orthocenter and is the midpoint of . On the line , take point such that . Show that . (hint: Use complex numbers – see here). Posted in Uncategorized || [...]