A hexagon inscribed in a circle

[Batalov]

Took a random problem from W.Wu's riddle site which used to be defunct at berkeley, but now is restored.

A hexagon with sides of length 2, 7, 2, 11, 7, 11 is inscribed in a circle. 
Find the radius of the circle.

Brute force solution is ugly, but that's the only one I managed.

The answer is nice, though.

I wonder if there's an elegant solution. I suspect that the elegant one is to remember a lemma about inscribed polygons (or make one up from scratch).

[davieddy]

4R[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup] + c[sup]2[/sup] + abc/R

[davieddy]

Can this result be generalized to an inscribed 2Nagon with pairs
of sides identical?

David

[Batalov]

[COLOR=black]4R[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup], for N=2, and then induction :ha-ha: [/COLOR]

[davieddy]

Originally Posted by Batalov

[COLOR=black]4R[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup], for N=2, and then induction :ha-ha: [/COLOR]

How do you imagine I arrived at my result in the first place?

Ans: Two pints and a bit of sleep.

[philmoore]

Didn't Mally post a problem like this a few years back?

Hint:
For a quadrilateral inscribed in a circle, opposite angles must add up to 180[SUP]o[/SUP]. Combine right angle trigonometry with the law of cosines to get a cubic formula for the diameter d, or the radius r if you prefer. In general, we would expect to have to use the cubic formula, but in this case, there is a positive rational root and two negative irrational roots, so the positive root is the solution. I am sure there must be an elementary solution not using trigonometry, but I don't see any way around needing to solve a cubic equation.

[davieddy]

Originally Posted by philmoore

Didn't Mally post a problem like this a few years back?

Hint:
For a quadrilateral inscribed in a circle, opposite angles must add up to 180[sup]o[/sup]. Combine right angle trigonometry with the law of cosines to get a cubic formula for the diameter d, or the radius r if you prefer. In general, we would expect to have to use the cubic formula, but in this case, there is a positive rational root and two negative irrational roots, so the positive root is the solution. I am sure there must be an elementary solution not using trigonometry, but I don't see any way around needing to solve a cubic equation.

Indeed he did.
I remember his succincness well

David

[Batalov]

I reposted verbatim from W.Wu after napkin getting to the 6th-degree equation which I found ugly -- and I did first search the forum for "hexagon" (even though of course hexagon is merely a decoration).

Should have looked for "inscribed".

[philmoore]

Originally Posted by philmoore

Didn't Mally post a problem like this a few years back?

I found it, although I had posted it rather than Mally:
http://www.mersenneforum.org/showthread.php?t=7045
Mally's discussion of cubic equations was quite entertaining!

By the way, the problem should easily generalize to an arbitrary number of chords inscribed in a circle, making an n-gon. Call the sides a, b, c, etc.

Since we have sin[SUP]-1[/SUP](a/d)+sin[SUP]-1[/SUP](b/d)+sin[SUP]-1[/SUP](c/d)+ ... = pi and d > max{a,b,c,...}, the left hand side is a decreasing function of d and should have a single positive solution in d. I don't see that the problem would be in general very tractable algebraically, though.

[davieddy]

http://www.mersenneforum.org/showpos...2&postcount=26

and the subsequent ones

David