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Please see the following diagram:

enter image description here This is an isosceles triangle.

Let the angles be as follow: green = G, red = R, blue = B, purple = P

G = pi - 2B

Now my question is.. Based on the value of blue angle (B)- is it possible to determine the R and P angles given the radius and arc length. The angles are formed by the tangent of the circles at the intersection point angle

Arc length is the "outside" arc length, for example, for the bottom right circle, it goes from point G to point H. Radius is simple-

All circles have the same radius, R1=R2=R3. As for the arc length, I believe the two bottom circles have the same L1=L2, but the upper circles's will be different.

I tried working it out, I obtained the value for the purple angle to be : 2pi - L/R; Can someone verify this as well? I am mostly interested in the red angle.

Thank you!

Koopa
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  • Can't you adapt my answer to your previous post to this situation? You know the interior angles, so you know the exterior ones as well, and with $\frac LR$ you also know how much is covered by the circle. The rest should be simple. – MvG Jul 15 '14 at 00:30
  • @MvG Can you help me with a diagram? your previous answer's was easy to read for me (I am a visual learner) but I am not very good at drawing just yet. I will try to figure out the rest myself! Thx – Koopa Jul 15 '14 at 21:42
  • It's not as easy as I thought at first: I incorrectly assumed that the angle bisectors as $A$ and $B$ would bisect the outer arc $L$, but that's not the case. Perhaps a reasonable first step would be working out how large the circles are with respect to the triangle. The ratio $L/R$ encodes that, but rather indirectly. The fractions $\frac a\alpha=\frac b\beta=\frac c\gamma$ are all equal due to the law of sines. If you could express that fraction in terms of $R$ and $L$, then you could build on that. – MvG Jul 15 '14 at 22:03
  • @MvG Thanks for your tips. Sadly, After trying for a few hours, I still have little idea how to do this. – Koopa Jul 16 '14 at 18:09

1 Answers1

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This is a complete rewrite of my previous post.

Assume $\tan B=h$. Up to a similarity transformation, the corners of your triangle will have the coordinates

$$A=\begin{pmatrix}-1\\0\end{pmatrix}\qquad B=\begin{pmatrix}1\\0\end{pmatrix}\qquad D=\begin{pmatrix}0\\h\end{pmatrix}$$

Now assume a circle radius $r$ and you can compute the points of intersection as

$$G=\begin{pmatrix} \frac{h^{2} + \sqrt{-h^{4} + 4 \, {\left(h^{2} + 1\right)} r^{2} - 2 \, h^{2} - 1} h + 1}{2 \, {\left(h^{2} + 1\right)}} \\[2ex] \frac{h^{3} + \sqrt{h^{2} + 1} \sqrt{-h^{2} + 4 \, r^{2} - 1} + h}{2 \, {\left(h^{2} + 1\right)}} \end{pmatrix}\qquad H=\begin{pmatrix}0 \\ -\sqrt{r^{2} - 1}\end{pmatrix}$$

Now the arc around point $B$ is

$$\frac{L_2}{R_2}=2\pi- \arccos\frac{(G-B)\cdot(H-B)}{\lVert G-B\rVert\cdot\lVert H-B\rVert}$$

Plugging in the above coordinates, you can obtain a relation between $r$ and $\frac{L_2}{R_2}$, but that fraction is highly non-linear. Therefore it is very difficult to compute $r$ given $\frac{L_2}{R_2}$. So it should be possible to solve this numerically for a given set of numbers, but I doubt I'll be able to come up with a general closed formula for the angles.

MvG
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  • thanks for the write-up, I will attempt this. This is a personal project of mine so it's no stress :) Have a good day! – Koopa Jul 16 '14 at 20:30
  • I hope this isn't too much to ask, but when you have free time could you attempt to solve it? I am not very good at this. I am very grateful for your efforts, thanks. – Koopa Jul 18 '14 at 22:39
  • @Koopa: Will be tricky. I've posted my results about as far as I get with them. – MvG Jul 21 '14 at 12:27
  • Thanks for your effort, I just realize now that the problem might be a bit too abstract. – Koopa Jul 21 '14 at 19:40
  • I've started to play around with the formula (plugging and the like in Wolfram Mathematica)- I have a silly question though, which part of it represents the angle G? Would it be what's inside the arccos term. Thanks for clarification – Koopa Jul 24 '14 at 20:58
  • @Koopa: For the angle $G$, since $G=\pi-2B$ you have $B=\frac{\pi-G}2$ so $h=\tan B=\tan\frac{\pi-G}2$. The $G$ inside the $\arccos$ term is the point $G$, computed a few lines above that. – MvG Jul 24 '14 at 23:11