Interesting math problem that I saw on-line

[Charles Link]

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.

Problem is contained in the summary above.

 

[Charles Link]

For a hint on this problem, opposite angles of the 4 sided polygon add to $\pi$ radians when inscribed in a circle.

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.

 

[sbrothy]

I was just about to protest that it sounded more like an ellipse, but on my first read I mistakenly had the polygon on the outside!

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

 

[Charles Link]

It does make sense because if you have 4 sticks/chords, you can always find a circle that is too large or too small. There is only one circle that works.

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 $\cos{\theta}$.

 

[sbrothy]

Man, I’m gonna use pen and paper for this one!

 

[sbrothy]

And…. I don’t have a ruler. This is gonna be a mess.

 

[Charles Link]

@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.

 

[sbrothy]

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.

 

[Charles Link]

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.

 

[Charles Link]

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

 

[Charles Link]

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. Then the first one has a diameter of about 5.25" (just under), and the second one would have a diameter of just over 5.5" if my arithmetic is correct.

 

[Charles Link]

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.

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

 

[Charles Link]

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.

 

[Charles Link]

I was just about to protest that it sounded more like an ellipse, but on my first read I mistakenly had the polygon on the outside!

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 mentioned as well.

 

[Charles Link]

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.

 

[Mayhem]

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.

I'm trying but it's not an easy problem.

 

[Mayhem]

Oh would you look at that. This is a solved problem. https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Parameshvara's_circumradius_formula

 

[Charles Link]

I'm trying but it's not an easy problem.

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 $\cos{\theta}$. Once you have that, you have 3 points from which to determine the center $(h,k)$. I'll write it out in the next post in a little while.

 

[mathwonk]

absurd "coincidence": when I "solved " it (completely wrong) by confusing tangent with cosine, I got sqrt(11/4); and the actual answer it seems is roughly 11/4. I apparently enjoy numerical coincidences, even nonsensical ones.

 

[Charles Link]

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 same in 3 cases.

 

[Charles Link]

Oh would you look at that. This is a solved problem. https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Parameshvara's_circumradius_formula

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.

 

[bob012345]

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 $$\pi =sin^{-1}(\frac{2}{2r})+sin^{-1}(\frac{3}{2r})+sin^{-1}(\frac{4}{2r})+sin^{-1}(\frac{5}{2r})$$

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.

 

[bob012345]

I guess they all do! No, never mind. There are conditions. One can’t even make a quadrilateral with just any four sides and some combinations will not be real in the formula.

 

[Charles Link]

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.

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.

 

[Mayhem]

Since math is a neverending problem, the question is whether a similar theorem exists / can be proven for a non-cyclic quadrilateral.

 

[Charles Link]

Since math is a neverending problem, the question is whether a similar theorem exists / can be proven for a non-cyclic quadrilateral.

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

 

[bob012345]

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.

Thanks! I managed to do it using my formula above as the starting point. Basically it involved untangling the trig functions into an algebraically solvable polynomial. It was probably more work than necessary.

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

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