Triangle tip following the tangent of a circle

Question

A maths problem i have here. I have a triangle that rotates in a circle, but i can't seem to wrap my head around, how to make it's head point in the direction of the tangent of the circle. I will put a picture of what i am expecting. May be i do not know how to properly formulate my question so feel free to edit it.

Any kind of insight will be helpful. Thank you!

Answer 1

I am going to assume that you want the triangle to point in the direction of the tangent at the head of the triangle (instead the tangent at the center of the triangle or some other point). I'll also assume that the triangle is isoscelese, and has base (the side opposite the head) of length $b$ and altitude (distance from head to base) of length $h$. If your triangle is equilateral with sidelength $k$, then $b = k$ and $h = \dfrac {\sqrt 3}{2}k$.

The tangent to a circle at a point $P$ is perpendicular to the radius of the circle at $p$. Assuming you are doing the traditional expression of a circle with center $(0,0)$ and radius $r$, you are probably already calculating the coordinates of $p$ as $(r\cos \theta, r\sin \theta)$. In this case, the direction of the radius is just the vector $(\cos \theta, \sin \theta)$. If the direction of the tangent is given $\hat t = (u, v)$, then the dot product with the radius direction should be $0$:

$$u\cos\theta + v\sin\theta = 0$$

Two obvious solutions to this are $(u,v) = (-\sin\theta, \cos\theta)$ and its opposite $(u,v) = (\sin \theta, -\cos\theta)$. These are the only two solutions that are unit vectors (i.e., their length is $1$). Your drawing indicates that you want it to point in the counter-clockwise direction, so $$\hat t = (-\sin\theta, \cos\theta)$$ is the tangent vector you want.

Now, the head of the triangle is at $p$. The midpoint $q$ of the base should be directly behind it at a distance of $h$ along the tangent. That is, it will be at $p - h\hat t$, or

$$q = (r\cos\theta, r\sin\theta) - h(-\sin\theta, \cos\theta) = (r\cos\theta + h\sin\theta, r\sin\theta - h \cos\theta)$$

The two base vertices are then at a distance $b/2$ from $q$ in either direction perpendicular to the tangent line. I.e., the direction of the radius. So these will be at $q \pm \frac b2(\cos\theta, \sin\theta)$.

Thus the three vertices of your triangle will be:

• Head: $(r\cos\theta, r\sin\theta)$
• Inner vertex: $\left(\left(r - \dfrac b2\right)\cos\theta + h\sin\theta, \left(r - \dfrac b2\right)\sin\theta - h \cos\theta\right)$
• outer vertex: $\left(\left(r + \dfrac b2\right)\cos\theta + h\sin\theta, \left(r + \dfrac b2\right)\sin\theta - h \cos\theta\right)$