## Thursday, November 24, 2005

### Simpler than it Sounds. Topic: Complex Numbers. Level: AMC/AIME.

Problem #1: Find the coordinates of the point $(5,4)$ rotated around the point $(2,1)$ by $\frac{\pi}{4}$ radians counterclockwise.

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

We want to rotate $5+4i$ around $2+i$. Rewrite these in $(r, \theta)$ form, where $r$ is the magnitude and $\theta$ is the angle. We also have $(r, \theta) = e^{i\theta}$ by the Euler Formula.

So, to simplify things, we shift $2+i$ to the origin, and now we want to rotate $3+3i = 3\sqrt{2}e^{i\frac{\pi}{4}}$. We notice that rotation by $\frac{\pi}{4}$ simply adds $\frac{\pi}{4}$ to the angle, resulting in $3\sqrt{2}e^{i\frac{\pi}{2}} = 3\sqrt{2}i$.

Shifting back to the "real" origin, our rotated point becomes $2+(1+3\sqrt{2})i$ in the complex plane and $(2, 1+3\sqrt{2})$ in the Cartesian plane. QED.

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Problem #2: Given that $A(x_1,y_1)$ and $B(x_2,y_2)$ are two vertices of an equilateral triangle, find all possible values for the third point, $C$.

Solution: Convert them to complex numbers - $X = x_1+y_1i$ and $Y = x_2+y_2i$.

Notice that if we rotate $X$ around $Y$ by $\frac{\pi}{3}$ or $-\frac{\pi}{3}$ radians, we get the only two possible points for $Z$ ($C$ as a complex number).

So applying the same rotation method as in the first problem, we have $Z = Y + (X-Y)e^{i\frac{\pi}{3}}$ or $Z = Y+(X-Y)e^{-i\frac{\pi}{3}}$. 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 $A = 3+i$ rotated by $\frac{\pi}{2}$ radians counterclockwise.

Practice Problem #2: (WOOT Class) Show that, given three complex numbers $A, B, C$ that lie on the unit circle, the orthocenter of the triangle formed by them is $H = A+B+C$ (hint: If two complex numbers $X, Y$ are perpendicular, $\frac{X}{Y}$ is purely imaginary... prove it).

Practice Problem #3: (WOOT Message Board) Let $O$ be the center of a circle $\omega$. Points $A,B,C,D,E,F$ on $\omega$ are chosen such that the triangles $OAB,OCD,OEF$ are equilateral. Let $L,M,N$ be the midpoints of $BC,DE,FA$, respectively. Prove that triangle $LMN$ is equilateral.

1. 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...

2. 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. Oops...so it's just multiplying by i? Then A' = -1 + 3i = (-1,3)?

4. [...] 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 || [...]