1. ## Interesting rotation problem

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 attempt at a solution:

At first, I thought that this problem might be solved by placing the whole thing on a coordinate axis... Not going to happen. There are so many arbitrary variables that even if I arrived at a solution that way, it would be an ugly algebraic mess. After some considerable time spent staring at my diagram I decided on one thing: if we take points O_a, O_b, and O_c, I believe they will always form an equilateral triangle. Thankfully, this fact need not be proven first--if we prove that point C moves in a circle around this O_c, then nothing else need be said.

A friend has suggested that I use vectors to solve this. I don't see the solution. The following is what I understand about the solution: in our diagram, we will place points A and B at their arbitrary locations on their circles. Then, I will rotate them to positions A' and B', an arbitrary angle $\phi$ counterclockwise. Then, I will want to show that point C', the third point on the equilateral triangle A'B'C' is the same C' as that point achieved by rotating point C about O_c by the angle $\phi$. This will suffice to solve the problem. My issue is precisely that I don't know how to show that above fact.
Help!

2. Hello, Mazerakham!

Edit: You're right, of course.
. . . .I had two different angles, but "forgot" when I posted.

I agree with you . . . Analytic Geometry is a very messy approach.
This is as far as I got . . .

The first circle has center $(x_a,y_a)$ and radius $r_a.$

Point $A$ has parametric equations: . $\begin{Bmatrix}x &=& x_a + r_a\cos\theta \\ y &=& y_a + r_a\sin\theta \end{Bmatrix}$

The other circle has center $(x_b,y_b)$ and radius $r_b.$

Point $B$ has parametric equations: . $\begin{Bmatrix}x &=& x_b + r_b\cos\phi \\ y &=& y_b + r_b\sin\phi \end{Bmatrix}$

We need distance $d = \overline{AB}\!:$

$d \:=\:\sqrt{[(x_b+r_b\cos\phi) - (x_a+r_a\cos\theta)]^2 + [(y_b+r_b\sin\phi) - (y_a + r_a\sin\theta)]^2}$

Since $\Delta ABC$ is equilateral: . $\overline{AB} \,=\,\overline{BC}\,=\,\overline{AC}$

Hence, we want point $C(x,y)$ so that: . $\overline{CA} \,=\,\overline{CB} \,=\, d$

And that's all we have to do . . . Right!

3. Haha, one little thing:

Points A and B may have different phase displacements!!!!

ie, they will start off at different angles,

So, for simplicity, we can say that point A will have no phase displacement, and B will have phase displacement $\phi$. Therefore, point A would have the same parametrization as you gave, but point B will actually be parametrized by:

$\begin{Bmatrix}x &=& x_b + r_b\cos \left( \theta - \phi \right) \\ y &=& y_b + r_b\sin \left( \theta - \phi \right) \end{Bmatrix}$

But anyway, the Greeks would have spat on us for using such an approach. This problem, while VERY tricky, is also simple. As Glenn Stevens from Boston University told my number theory class,

$simple \neq easy$

I've spent hours and hours on this problem. Call me stubborn, but there has to be a better way than analytic geometry. Even if we place everything in "standard position" with the origin, I count at least 5 variables, and they just won't go away unless we try to solve this problem more creatively.

For those who are joining in, I'll tell you about the origin of this problem. It's from the All-Soviet-Union Mathematics Competition which ran from (I believe) 1961-1985. This problem was meant to be solvable by school-age kids (high school, I hope ).

I'll be working on it. I'll post anything interesting I figure out.

4. Vectors are the way to go. Vectors and rotation matrices. One of the key facts about rotation matrices is that in 2 dimension, 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.

5. 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 y will be different.

6. Yep, absolutely right about the radius multiplier, and of course it still works.

The reason I knew algebra would not be necessary for this problem was that, in fact, this problem works for ANY triangle--it didn't have to be an equilateral triangle. It could have been a 45-90-45 right triangle, where, say, point C forms the 90 degree vertex. So, it made sense to use some more... simple, fundamental ... concepts.

This was fantastic, once again Adrian. Can't thank you enough.

Jake

7. You're very welcome. A fun problem! If you have more of those, please do send them along.