Recent content by Charles Link

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} ##...

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...

Working it with the polygon on the outside I think is much more difficult, but you will also find a circle that works, if I'm not mistaken. Having it with the circle on the inside looks almost too difficult to readily get an answer. Perhaps someone will find a way to do this other case you...

The lack of response on this is disappointing. I thought it was one of the better math problems I had seen in the last couple of months. It's slightly difficult, but once you see how to solve it, it is fairly straightforward.

The law of cosines is basically the Pythagorean theorem when applied to triangles that don't have the 90 degree angle.

I might as well give the (approximate) answers to both of these, so that anyone who likes doing artwork can draw them up The first turns out to have ## r \approx 2.6 ## and the second has ## r=7 ##. It might make things easier to draw the first one in inches and the second one in centimeters...

We apparently must have a small number of viewers. I'm surprised nobody solved these two cases yet and posted the results.

@Andy Resnick It really came as a surprise to me. Where was Jupiter at 4 am? It should have been visible very near the moon, from what I observed the night/morning before. Instead Jupiter was behind the moon. :)

See https://www.physicsforums.com/threads/trigonometry-problem-of-interest.1082657/ The second case I gave comes from this problem, where when you compute it, you will find the diameter of the circle is the same as the largest chord.

@sbrothy Use the law of cosines to find one of the angles, and once you have that, you have 3 points to determine the circle. You put the center at ## (h,k) ## with the one point in the middle as the origin, and you can find ##(h,k) ## and thereby the radius.

The moon was a little above Jupiter the previous morning when it rose about 3:30 am, so I was expecting to see Jupiter near the moon this morning October 6 when the moon rose about 4 am. Even though the skies were fairly clear, Jupiter was not present. I was a little surprised, bit about 20...