Is Point C Moving in a Circle with Same Angular Speed and Direction?

[Ackbach]

Here is this week's POTW:

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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. Points $A$ and $B$ begin rotating in the same direction (say, counterclockwise) around their respective circles with the same angular speed (say, $\omega$) 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 ($\omega$) about some center $O_c$ somewhere in the plane.

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[Ackbach]

I am in the process of grading one submission; in the meantime, here is my solution:

Spoiler

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
$$\vec{A}=R_{\omega t}\,(r_a \hat{x}),$$ and the vector from the origin to point B be
$$\vec{B}=\vec{O}_{b}+R_{\omega t+\theta}\,(r_b \hat{x}),$$
for some (phase) angle $\theta$.
We want to show that
$$\vec{C}=\vec{O}_{c}+R_{\omega t}\,\vec{y},$$ for some constant vector $\vec{y}.$
Note that
$$\vec{AB}=\vec{B}-\vec{A}=\vec{O}_{b}+R_{\omega t+\theta}\,(r_b \hat{x})-R_{\omega t}\,(r_a \hat{x}).$$
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:
\begin{align*}
\vec{C}&=R_{\omega t}\,(r_a \hat{x})+R_{\pi/3}\left(\vec{O}_{b}+R_{\omega t+\theta}\,(r_b \hat{x})-R_{\omega t}\,(r_a\hat{x})\right) \\
&=R_{\pi/3}\vec{O}_b+R_{\omega t}\,\left[r_a \hat{x}+R_{\theta+\pi/3}\,(r_b \hat{x})-R_{\pi/3} \, (r_a \hat{x}) \right].
\end{align*}
So, let
$$\vec{O}_{c}=R_{\pi/3}\vec{O}_{b}$$ and
$$\vec{y}=r_a \hat{x}+R_{\theta+\pi/3}\,(r_b \hat{x})-R_{\pi/3} \, (r_a \hat{x}),$$
and you're done. QED.

 

[Ackbach]

Update: MarkFL also correctly answered this week's POTW. Here is his solution:

Spoiler

Let us orient our coordinate axes such that the origin is at $O_a$ and the centers of the two circles are on the $x$-axis, where $O_b$ is on the positive $x$-axis, at $(O_B,0)$. According to the problem, we may define points $A$ and $B$ parametrically with:

Point $A$:

\(\displaystyle x_A=r_a\cos(\omega t+\alpha)\)

\(\displaystyle y_A=r_a\sin(\omega t+\alpha)\)

Point $B$:

\(\displaystyle x_B=r_b\cos(\omega t+\beta)+O_B\)

\(\displaystyle y_B=r_b\sin(\omega t+\beta)\)

Now, point $C$ will lie along the line $y_c$ perpendicular to $\overline{AB}$ and passing through the mid-point of $A$ and $B$, and a distance of \(\displaystyle \frac{\sqrt{3}}{2}\overline{AB}\) from this mid-point.

Hence, we find the line $y_c$ to be:

\(\displaystyle y_c=-\frac{x_B-x_A}{y_B-y_A}\left(x-\frac{x_A+x_B}{2}\right)+\frac{y_A+y_B}{2}\)

And then the distance formula yields:

\(\displaystyle \left(x_C-\frac{x_A+x_B}{2}\right)^2+\left(y_C-\frac{y_A+y_B}{2}\right)^2=\frac{3}{4}\left(\left(x_B-x_A\right)^2+\left(y_B-y_A\right)^2\right)\)

Using the line, and multiplying through by 4, we get:

\(\displaystyle \left(2x_C-x_A-x_B\right)^2+\left(\frac{x_B-x_A}{y_B-y_A}\left(2x_C-x_A-x_B\right)\right)^2=3\left(\left(x_B-x_A\right)^2+\left(y_B-y_A\right)^2\right)\)

\(\displaystyle \left(2x_C-x_A-x_B\right)^2\left(\frac{\left(x_B-x_A\right)^2+\left(y_B-y_A\right)^2}{\left(y_B-y_A\right)^2}\right)=3\left(\left(x_B-x_A\right)^2+\left(y_B-y_A\right)^2\right)\)

\(\displaystyle \left(2x_C-x_A-x_B\right)^2=3\left(y_B-y_A\right)^2\)

And so we find:

\(\displaystyle x_C=\frac{\left(x_A+x_B\right)\pm\sqrt{3}\left(y_B-y_A\right)}{2}\)

\(\displaystyle y_C=\frac{\left(y_A+y_B\right)\mp\sqrt{3}\left(x_B-x_A\right)}{2}\)

Given the fact that the sum/difference of two sinusoids having the same period is itself another sinusoid having the same period as the two addends, along with the fact that the phase difference of the two coordinates is the same as for points $A$ and $B$, we have shown that the locus of point $C$ is a circle and its angular velocity is $\omega$.