Recent content by Charles Link

I was in a similar position as you 45 years ago, and I made a successful transition to the defense industry where we did a lot of applied research. In hindsight, the university professors in the solid state physics department were really spinning their wheels researching things like magnetic...

Using the formula of post 22 for ## r ##, I found using ## a^2+d^2=b^2+c^2 ## that the algebra is non-trivial to show that ## 4r^2=a^2+d^2=b^2+c^2 ##, but I got it to work. Perhaps there is a shortcut to the algebra. Of course there is also Thales' theorem that it had to be the case.

Notice when ## a+b+c-d ## is small, when the 4 sided polygon is very flat, that ## r ## is very large. The center of the circle is then very far from the polygon and is far outside of it. (This is just one example. The 3 other terms in the denominator all have a similar effect. Here I...

By ## d ## you are referring to the 4th side, and not the diameter of the circle. That took me a second look to figure out what you were referring to.

@anuttarasammyak I don't know that your method (referring to post 39) is valid. The ## x's ## are not variables with any freedom. You only need one expression, not 3 of them to solve for ## d ##. Edit: Meanwhile @bob012345 worked through all the difficult algebra on this one successfully...

This is from post 35 above. I don't know why ## \sin(a+b) ## doesn't seem to work on @bob012345 's formula. That's what I tried around post 28 and I found it to be very messy. In any case super job @bob012345 and also @kuruman and also @anuttarasammyak for some very healthy responses. :)

@kuruman Excellent. :) You got ## \cos{\theta} ## how I did, but once you have 3 points on the circle, (once you find ## \cos{\theta} ## ), you can compute center ## (h,k) ## and thereby you have ## r ##. Excellent solution you have in any case. :)

This is @bob012345 's approach, and he got the algebra to work above, basically getting a closed form solution for ## r ##. In his latest post, (#27), he's only a step or two away from getting the general solution for ## r ## in closed form.

It's a lot of extra algebra. I did it for the solution with ## \cos{\theta} ## that I solved above, and without a doubt, it will work for your solution as well. You already got through the most difficult part of the algebra. :)

@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...

The answer is no. If it is not contained in a circle, the 3rd angle can be chosen arbitrarily.

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...

Makes it too easy. It is worth the effort to derive these results. The math skills of this latest generation seem to be falling apart.

Try again. ## \cos(\pi-\theta)=-\cos{\theta} ##. Once you have ## \cos{\theta} ##, you have 3 points: ## (0,0) ##, ##(2,0) ##, and ## (3 \cos{\theta}, 3 \sin{\theta}) ##. You put in center of circle ## (h,k) ## and find ## h=1 ## and ## k ## is also easy to compute, setting the ## r^2 ##...

The two opposite angles adding to ## \pi ## radians (when a 4 sided polygon is inscribed in a circle) makes it readily workable. I may post the solution shortly. Observe: Law of cosines: ## L^2=2^2+3^2-2(2)(3) \cos{\theta}=4^2+5^2-2(4)(5) \cos(\pi-\theta) ##. Solve for ## \cos{\theta} ##...