Place 12 equidistant dots on a $\pi$ / 3 curve

For a computer game (which means the origin is on the top left) I need to place 12 dots equidistantly on a circle.

The curve should also go through the 3 red dots shown below:

Here is how I am trying to solve it:

The dots on the curve must have the coordinates:

x = R * cos(alpha + beta) + p
y = R * sin(alpha + beta) + q


The R, beta, p and q are unknown to me, but alpha goes from 0 to $\pi$/3 in 12 steps and I know w, h and b.

And I know that the curve should contain the 3 red dots:

(0, h - 2b) when alpha is 0
(w/2, h - 3b) when alpha is pi/6
(w, h - 2b) when alpha is pi/3


So this gives me 6 equations to solve:

0 = R cos(beta) + p
h - 2b = R sin(beta) + q

w/2 = R cos(pi/6 + beta) + p
h - 3b = R sin(pi/6 + beta) + q

w = R cos(pi/3 + beta) + p
h - 2b = R sin(pi/3 + beta) + q


The above equations should provide me with R, beta, p and q - right?

And then I could place my 12 dots (they are actually midpoints of playing cards) by running this loop:

for (i = 0; i < 12; i++) {
card.x = R * Math.cos(i * Math.PI / 36) + p;
card.y = R * Math.sin(i * Math.PI / 36) + q;
}


Unfortunately, I don't know how to solve the 6 equations above (except that p is probably w/2).

Below is the current screenshot of my card game, where I am trying to place the bottom cards on a curve (around the player avatar):

UPDATE: I've given up the curve for the moment and tried to solve the equation

y = px^2 + qx + r


for the 3 dots and got:

p = 4b / w^2
q = - pw
r = h - 2b


This gives me quite nice card placement:

But if someone posts a complete solution for the curve, that would be great.

My math skills are too rusty and I probably do some basic errors while solving...

-
so would you prefer the cards to be on a circular or on a parabolic curve? –  long tom Mar 25 '13 at 9:29
Actually they look good enough now on a parabolic curve, but I am curious about the solution to the "real" curve too. –  Alexander Farber Mar 25 '13 at 10:07

I'd try to find the radius of the circle by solving $r^2=(r-b)^2+(w/2)^2$.
The angle $\gamma$ between the radii to the outer spots (on the border of the screen) should be the solution of $\sin \gamma/2=\dfrac{w/2}{r}$.
(indeed $p=w/2$, $q=h+r-3b$)
The angle between the radii to the outer spot - can't I deliberately set it to $\pi$/3? (i.e. I wonder: how many circles can go through 3 dots?) –  Alexander Farber Mar 23 '13 at 19:35
@Alexander: If you want to keep the angle to $\pi/3$, you need to change those $2b$, $3b$ numbers. –  Jyrki Lahtonen Mar 23 '13 at 19:53
If you want to keep the angle to $\pi/3$ and keep $2b$ (but not $3b$), then $r=w$ –  long tom Mar 23 '13 at 20:17