A hexagon inscribed in a circle
 

Took a random problem from W.Wu's riddle site which used to be defunct at berkeley, but now is restored.
[code]A hexagon with sides of length 2, 7, 2, 11, 7, 11 is inscribed in a circle.
Find the radius of the circle. [/code]

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

[spoiler]The answer is nice, though.[/spoiler]

[spoiler]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).[/spoiler]

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

Followup puzzle
 

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

David

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

[quote=Batalov;217132]:smile: [COLOR=black][spoiler][COLOR=black]4R[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup], for N=2, and then induction :ha-ha: [/COLOR][/spoiler][/COLOR][/quote]

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

Ans: [spoiler]Two pints and a bit of sleep.[/spoiler]

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

Hint:
[SPOILER]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.[/SPOILER]

[quote=philmoore;217140]Didn't Mally post a problem like this a few years back?

Hint:
[spoiler]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.[/spoiler][/quote]
Indeed he did.
I remember his succincness well:smile:

David

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". :doh!:

[QUOTE=philmoore;217140]Didn't Mally post a problem like this a few years back?[/QUOTE]

I found it, although I had posted it rather than Mally:
[url]http://www.mersenneforum.org/showthread.php?t=7045[/url]
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.

[SPOILER]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.[/SPOILER]

You, me and the late lamented
 

[URL]http://www.mersenneforum.org/showpost.php?p=97752&postcount=26[/URL]

and the subsequent ones:smile:

David