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I have the general situation illustrated below. I know angles $\Theta_{1}, \Theta_{2}$ and line lengths $A$, $B$ and $C$, (where $A = a_{1} + a_{2}$ etc.). I'm trying to calculate the largest semi-circle I can fit on line $B$ (i.e., calculate r), that will (always?) be tangent to line $A$ and $C$ when the angles are acute.

General Situation

So I think: $$ r = \frac{B.sin\Theta_{1} .sin\Theta_{2} } { sin\Theta_{1} + sin\Theta_{2} } $$

For the cases where $A > a_{1}$ and $C > c_{1}$ (meaning A is longer than the theoretical tangent point at the angle $\Theta_{1}$ etc.). But what if $A$ was very short? Does $r$ simply become the distance to the end point of $A$? And how do I 'know' when this is the case mathematically?

EDIT: To add some clarification...

  • This is part of a larger effort to find the largest semi-circle that will fit inside an arbitrary closed polygon.
  • The lines illustrated are thus three consecutive segments of that polygon
  • They may be any angle and length, and don't form a triangle unless my polygon is a triangle.
  • Clearly, if the angle at either or both ends is larger than $\pi$, the semi-circle can go right to that vertex and isn't impinged upon by the next or previous segment.

The part I'm mostly struggling with is incorporating the line length in determining $r$ and the centre point. Obviously, it has an effect at any given angle:

Lines

My ultimate idea is to draw lines between every polygon vertex and calculate the largest semi-circle that will fit inside the polygon along that line. I hope that will give me a good guess at the largest generally that will fit.

EDIT 2: Specifics... Specifc

I know the blue values, how do I calculate $r$ (and thus $b_{2}$ and $b_{3}$)? ($h$ and $b_{1}$ being trivial). It seems like it should be easy...but I just can't seem to get my head around it.

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  • $\begingroup$ I had answered your question but withdrew it when I realised you were not necessarily forming a triangle with the three line segments. Which brings me to a question: does the semi-circle have to touch line segments $A$ and $B$, or be tangents of lines $A$ and $B$ (in other words imagine lines $A$ and $B$ extending to infinity)? $\endgroup$ – robert timmer-arends Mar 20 '20 at 8:59
  • $\begingroup$ I was able to prove that the expression in the original question is correct. I can provide the proof if necessary. From the proof, by extension, it follows that for $\theta_1, \theta_2 < \pi/2$ the point is always on the line segment $B$, and the tangents are always on the line segments $A$, $C$ $\endgroup$ – Aleksejs Fomins Mar 20 '20 at 9:05
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    $\begingroup$ (Changing side names to lower-case ...) The "largest" semi-circle on side $b$ that will always be tangent to (the lines of) sides $a$ and $b$ is in fact the only such semi-circle. (Your formula for the radius is correct.) If $a$ is "very short" and cannot reach the "theoretical tangent point" of that semi-circle, then no semi-circle will do. (Likewise for $c$.) So, you must have $$a \geq a_1 = r \frac{\cos\theta_1}{\sin\theta_1} = b\frac{\cos\theta_1\sin\theta_2}{\sin\theta_1+\sin\theta_2} \quad\text{and}\quad c \geq c_1 = b\frac{\sin\theta_1\cos\theta_2}{\sin\theta_1+\sin\theta_2}$$ $\endgroup$ – Blue Mar 20 '20 at 9:17
  • $\begingroup$ Are the ends of the segments A and C arbitrary? It would make more sense to have them end where they meet so you wouldn't have this problem. In any case your solution is always correct since it only depends on B and the two angles. $\endgroup$ – Sophie Mar 20 '20 at 9:30
  • $\begingroup$ I expanded on the problem a bit to explain further what I'm attempting to do. $\endgroup$ – Kyudos Mar 22 '20 at 20:46
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enter image description here

In the figure, $P'$ and $Q'$ are projections of $P$ and $Q$ onto $\overline{AC}$; and $P''$ is the reflection of $P$ in that segment. $M$ is the midpoint of $\overline{P''Q}$, and $S$ completes the rectangle $\square P'Q'QS$.

By the Inscribed Angle Theorem, we're guaranteed that $\angle QPP''\cong\angle P''KM$, so that the sine of that angle can be written in two ways:

$$\frac{|QS|}{|PQ|} = \frac{|P''M|}{r} \quad\to\quad r = \frac{|PQ||P''M|}{|QS|} = \frac{|PQ|\cdot\frac12|P''Q|}{|P'Q'|} = \frac{|PQ||P''Q|}{2\,|P'Q'|} \tag{$\star$}$$

In terms of provided parameters, we have $$\begin{align} |P'Q'| &= |AC|-(|AP'|+|Q'C|) \\[4pt] &= b - \left(a \cos\theta_1 + c\cos\theta_2\right) \tag{1}\\[6pt] |PQ|^2 &= |P'Q'|^2 + |PS|^2 = |P'Q'|^2 + (|PP'|-|QQ'|)^2 \\[4pt] &= |P'Q'|^2 + \left(a \sin\theta_1-c\sin\theta_2\right)^2 \tag{2}\\[6pt] |P''Q|^2 &= |P'Q'|^2 + |P''S|^2 = |P'Q'|^2 + (|PP'|+|QQ'|)^2 \\[4pt] &= |P'Q'|^2 + \left(a \sin\theta_1+c\sin\theta_2\right)^2 \tag{3} \end{align}$$

Now, one readily shows that

Point $P$, respectively $Q$, is a point of tangency when $$ a = b\;\frac{\cos\theta_1\sin\theta_2}{\sin\theta_1+\sin\theta_2} , \quad\text{resp.}\quad c = b\;\frac{\sin\theta_1\cos\theta_2}{\sin\theta_1+\sin\theta_2} \tag{$\star\star$} $$

For computational purposes in $(\star)$, lengths $a$ and $c$ should be taken no larger than the values given in $(\star\star)$.

Here are some examples of the results:

enter image description here

enter image description here

enter image description here

enter image description here

The last two have the same semicircle, since it was computed from the same maximal values of $a$ and $c$ from $(5)$.

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  • $\begingroup$ I haven't got back to working on this, but out of interest - what did you use to make these diagrams? $\endgroup$ – Kyudos Sep 14 '20 at 1:09
  • $\begingroup$ @Kyudos: I use GeoGebra for my diagrams. $\endgroup$ – Blue Sep 14 '20 at 1:14
  • $\begingroup$ Looking at this further - if a and b are both shorter than the tangential points, P and Q are on the circle and the centre is on line AC and thus I can find r. However, how to solve if only one line (say AP) is shorter than its tangential point? P is on the circle, the centre is on AC but there will be a different tangential point on CQ? $\endgroup$ – Kyudos Feb 10 at 2:12
  • $\begingroup$ @Kyudos: Equation $(\star)$ gives the radius in all viable cases. Viability is determined by whether $a$ and $c$ are no larger than the values calculated in $(\star\star)$. $\endgroup$ – Blue Feb 10 at 2:35
  • $\begingroup$ Thanks - think I'm getting to grips with it! Thanks for the GeoGebra tip too...really cool and useful program! $\endgroup$ – Kyudos Feb 10 at 3:11
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Based on your edit 2:

$b_1=a\cdot cos(\alpha)$ and $r=b_3\cdot sin(\beta)$. Hence $r^2=b_3^2\cdot sin^2(\beta)$

Let $k=b_1+b_2$. Then, from the Cosine Rule: $r^2=a^2+k^2-2\cdot a\cdot k\cdot cos(\alpha)$

Equating the two expressions for $r^2$, and collecting terms around $k$, gives $$cos^2(\beta)\cdot k^2-2(a\cdot cos(\alpha)-b\cdot sin^2(\beta))\cdot k+a^2-b^2\cdot sin^2(\beta)=0$$

which can be solved for $k$ using the quadratic formula (it gets quite cumbersome!). Of course this gives two values, but in the example I tried, one could be easily eliminated since it was negative.

Having found $k=b_1+b_2$, $b_3$ can be readily calculated, and hence $r$.

As indicated above, I only tested this once, setting up the situation in Geogebra, measuring the angles, and drawing a circle to fit. The calculated radius and measured radius were in agreement, so hopefully this is of some use.

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  • $\begingroup$ Thanks Robert, I set this up in GeoGebra and it works out. $\endgroup$ – Kyudos Feb 10 at 20:40
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It looks like the line from the intersection of sides A and C is the angle bisector because of the two perpendiculars of length r.

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