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.


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.


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. 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 || [...]