TL;DR

You have a 4 sided polygon inscribed in a circle. Chord lengths are 2,3,4,5 in any order. Find the radius of the circle. It is also of interest to work the same thing for chords of length 2,7, 11, and 14.

Edit: I'm going to add to this, that if some people are stuck on it, that I didn't solve it instantaneously either. I was able to solve it though.

EDIT: and now I’m not even sure if that makes any sense at all….

I'm going to add another hint or two: Use the law of cosines twice for two opposite angles. The ## L^2 ## on the side opposite this angle is the same for both. Solve for ## \cos{\theta} ##.

I know I have enough clues by now. Something tells me the solution relates to Pythagoras. It’s simple geometry in fact. I have a bunch of excuses though. A laptop gone up in flames, literally. And I’ve just eaten. Homemade lasagne with lots of of cheddar. So I might have to pass today.

and it is disappointing that no one else seems to be solving this. In the post that I got this from on-line, a fair number of people solved it before I did. Physicsforums doesn't seem to have the audience right now.

I even solved the problem now in the OP, expressing the ## \cos{\theta} ## in terms of the sides of the polygon, and getting an answer in terms of ## a,b,c, ## and ##d ## that is symmetric with them. I checked my answer for the 2,3,4,5 case, and I got the same answer: ## r=\sqrt{986.7}/ 12 \approx 2.6 ##.

and it is disappointing that no one else seems to be solving this. In the post that I got this from on-line, a fair number of people solved it before I did. Physicsforums doesn't seem to have the audience right now.

$$r=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$

It might be interesting to study what the limits of parameters have real solutions.

I did not see this problem until last night but I solved it at least numerically by using the chord formula and knowing the sum of the half angles is ##\pi##. I got $$\pi =sin^{-1}(\frac{2}{2r})+sin^{-1}(\frac{3}{2r})+sin^{-1}(\frac{4}{2r})+sin^{-1}(\frac{5}{2r})$$

View attachment 374739 Regarding Parameshvara’s equation mentioned above for the radius of the circle which inscribed the quadrilateral where ##s=\frac{a+b+c+d}{2}##

$$r=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$

It might be interesting to study what the limits of parameters have real solutions.

Meanwhile, thank you @bob012345 =Bob V., you have a very interesting way of working it. You don't readily get a closed form for ## r ## with your method, but I think it still might be possible to get it with a little extra work. I was able to get such a closed form with the method I used, and it agreed with the formula that @Mayhem linked above.

For real solutions, the longest chord can't be larger than the sum of the other three.

Meanwhile, thank you @bob012345 =Bob V., you have a very interesting way of working it. You don't readily get a closed form for ## r ## with your method, but I think it still might be possible to get it with a little extra work. I was able to get such a closed form with the method I used, and it agreed with the formula that @Mayhem linked above.

Let

##x=\frac{1}{r}.##

We start with the chord formula and angle sum in post #22;

##\arcsin\left(\frac{x}{2}\right)+\arcsin\left(\frac{3x}{2}\right)+\arcsin(2x)+\arcsin\left(\frac{5x}{2}\right)=\pi.##

Rearranging,

##\arcsin(x)+\arcsin\left(\frac{3x}{2}\right)=\pi-\left[\arcsin(2x)+\arcsin\left(\frac{5x}{2}\right)\right].##

Taking the cosine of both sides gives

##\begin{aligned}
&\cos\left(\arcsin(x)+\arcsin\left(\frac{3x}{2}\right)\right)\
=-\cos\left(\arcsin(2x)+\arcsin\left(\frac{5x}{2}\right)\right).
\end{aligned}##

Using the cosine addition formula,

$$\begin{aligned}
&\cos(\arcsin x)\cos\left(\arcsin\frac{3x}{2}\right)
-\sin(\arcsin x)\sin\left(\arcsin\frac{3x}{2}\right)\
&=-\left[
\cos(\arcsin 2x)\cos\left(\arcsin\frac{5x}{2}\right)
-\sin(\arcsin 2x)\sin\left(\arcsin\frac{5x}{2}\right)
\right].
\end{aligned}$$

Since

##\sin(\arcsin x)=x,\qquad \cos(\arcsin x)=\sqrt{1-x^2},##

we obtain

##\begin{aligned}
&\sqrt{1-x^2}\sqrt{1-\frac{9x^2}{4}}-\frac{3x^2}{2}\
=-\left(\sqrt{1-4x^2}\sqrt{1-\frac{25x^2}{4}}-5x^2\right).
\end{aligned}##

Equivalently,

##\begin{aligned}
&\sqrt{1-\frac{13x^2}{4}+\frac{9x^4}{4}}
+\sqrt{1-\frac{41x^2}{4}+25x^4}\
=\frac{13x^2}{2}.
\end{aligned}##

Squaring twice and simplifying yields

##x^2(3289x^2-480)=0.##

Discarding the zero solution, we find

##x^2=\frac{480}{3289},\qquad x=\sqrt{\frac{480}{3289}}.##

Since ##x=1/r,## we obtain

##\boxed{r=\sqrt{\frac{3289}{480}}\approx 2.61765}.##

Edit: But I see your post above, it looks like you are very close to being able to do precisely that. :) Excellent job. :)

@bob012345 You definitely get the correct numerical answer with your method, but I had no success in using your formula to get an answer for ## r ## in closed form in terms of ## a,b,c ## and ## d ##.

Edit: But I see your post above, it looks like you are very close to being able to do precisely that. :) Excellent job. :)