Favorite Old Threads - Best Math Thread # 3

[Ackbach]

Originally posted by Mazerakham, August 9th, 2010.

This problem is a doozy:

A circle is in the plane with center at O_a and some radius r_a. Another circle, not touching the first circle anywhere, has center at O_b and some radius r_b. Points A and B are both (arbitrary) points on circle A and circle B respectively. Finally, Point C is in such a location that ABC is an equilateral triangle. Suddenly, points A and B begin rotating in the same direction (say, counterclockwise) around their respective circles with the same angular speed (say, w) about their centers. During this process, point C moves so that ABC remains an equilateral triangle.

Prove that point C is moving in a circle with same direction (counterclockwise) and angular speed (w) about some center O_c somewhere in the plane.

My solution:Vectors are the way to go. Vectors and rotation matrices. One of the key facts about rotation matrices is that in 2 dimensions, anyway, two rotation matrices commute.

Let $\hat{x}=\begin{bmatrix}1\\0\end{bmatrix}$ be the unit vector in the $x$ direction, and let
$$R_{\theta}=\begin{bmatrix}\cos(\theta) &-\sin(\theta)\\
\sin(\theta) &\cos(\theta)\end{bmatrix}$$ be the rotation matrix through $\theta$ radians.
Fact:
$$R_{\varphi}R_{\theta}=R_{\varphi+\theta}=R_{ \theta+\varphi}=R_{\theta}R_{\varphi}.$$
Without loss of generality, we may let the vector from the origin to point A be
$$\vec{A}=R_{\omega t}\,\hat{x},$$ and the vector from the origin to point B be
$$\vec{B}=\vec{O}_{b}+R_{\omega t+\theta}\,\hat{x}.$$
We want to show that
$$\vec{C}=\vec{O}_{c}+R_{\omega t}\,\vec{y},$$ for some constant vector $\vec{y}.$
Note that
$$\vec{AB}=\vec{B}-\vec{A}=\vec{O}_{b}+R_{\omega t+\theta}\,\hat{x}-R_{\omega t}\,\hat{x}.$$
Also note that
$$\vec{AC}=R_{\pi/3}\,\vec{AB},$$ by virtue of ABC being an equilateral triangle. Finally, we see that
$$\vec{C}=\vec{A}+\vec{AC}.$$ Thus, we compute:
$$\vec{C}=R_{\omega t}\,\hat{x}+R_{\pi/3}\left(\vec{O}_{b}+R_{\omega t+\theta}\,\hat{x}-R_{\omega t}\,\hat{x}\right)$$
$$=R_{\pi/3}\,\vec{O}_{b}+\left(R_{\omega t}+R_{\pi/3}R_{\omega t}R_{\theta}-R_{\pi/3}R_{\omega t}\right)\hat{x}$$
$$=R_{\pi/3}\vec{O}_{b}+R_{\omega t}\left(I+R_{\pi/3}R_{\theta}-R_{\pi/3}\right)\hat{x}.$$
So, let
$$\vec{O}_{c}=R_{\pi/3}\vec{O}_{b}$$ and
$$\vec{y}=\left(I+R_{\pi/3}R_{\theta}-R_{\pi/3}\right)\hat{x},$$
and you're done. QED.

Second Post:

Whoops, I've got a slight (fixable) error. You need to put in multipliers for the radii times the unit vector. But if you carry that through, you'll still be able to factor out the critical rotation matrix. Your vector $\vec{y}$ will be different.

 

[lofgran]

My revised solution:

Vectors are the way to go. Vectors and rotation matrices. One of the key facts about rotation matrices is that in 2 dimensions, anyway, two rotation matrices commute.

Let $\hat{x}=\begin{bmatrix}1\\0\end{bmatrix}$ be the unit vector in the $x$ direction, and let
$$R_{\theta}=\begin{bmatrix}\cos(\theta) &-\sin(\theta)\\
\sin(\theta) &\cos(\theta)\end{bmatrix}$$ be the rotation matrix through $\theta$ radians.
Fact:
$$R_{\varphi}R_{\theta}=R_{\varphi+\theta}=R_{ \theta+\varphi}=R_{\theta}R_{\varphi}.$$
Without loss of generality, we may let the vector from the origin to point A be
$$\vec{A}=r_aR_{\omega t}\,\hat{x},$$ and the vector from the origin to point B be
$$\vec{B}=r_b(\vec{O}_{b}+R_{\omega t+\theta}\,\hat{x}).$$
We want to show that
$$\vec{C}=r_c(\vec{O}_{c}+R_{\omega t}\,\vec{y}),$$ for some constant $r_c$ and vector $\vec{y}.$
Note that
$$\vec{AB}=\vec{B}-\vec{A}=r_b(\vec{O}_{b}+R_{\omega t+\theta}\,\hat{x})-r_aR_{\omega t}\,\hat{x}.$$
Also note that
$$\vec{AC}=r_cR_{\pi/3}\,\vec{AB},$$ by virtue of ABC being an equilateral triangle. Finally, we see that
$$\vec{C}=r_aR_{\omega t}\,\hat{x}+r_cR_{\pi/3}\left(\vec{O}_{b}+R_{\omega t+\theta}\,\hat{x}-r_aR_{\omega t}\,\hat{x}\right)$$
$$=r_aR_{\omega t}\,\hat{x}+r_cR_{\pi/3}\vec{O}_{b}+r_cR_{\pi/3