
New proof of a minimum property of the regular n-gon (1947) - mgdo
http://fermatslibrary.com/s/new-proof-of-a-minimum-property-of-the-regular-n-gon
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cousin_it
In other words:

1) Take a regular n-gon P.

2) Inscribe a circle C1, and circumscribe a circle C2.

3) The parts of C2 that lie outside P are n circle segments, whose straight
sides are tangent to C1.

4) If you move these segments around in any way, they will overlap each other,
so the uncovered area of C2 will become larger.

5) Therefore P is the smallest possible n-gon whose inscribed circle is C1.

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SilasX
So the new property found is that, for any given circle C1, the smallest n-gon
whose inscribed circle is C1 is a regular n-gon?

Edit: rephrasing.

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GFK_of_xmaspast
Needs a (1947).

Full cite: Tóth, L. Fejes. "New Proof of a Minimum Property of the Regular
n-GON." The American Mathematical Monthly 54.10 (1947): 589-589.

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kens
Is there any reason why this proof from 1947 is on the front page? Is this
proof particularly gratifying of intellectual curiosity or is this just a
random feedback loop?

~~~
Someone
It is the "paper of the week" on
[http://fermatslibrary.com](http://fermatslibrary.com), so quite a few people
will have seen it for the first time recently.

Being on that site typically means that the proof is short and reasonably
accessible for amateurs (you won't see "A monad is just a monoid in the
category of endofunctors" style proofs that aren't hard, but require one to
grasp terminology that few master there)

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ccvannorman
As an amateur math enthusiast, I find this kind of thing both fascinating and
frustrating. It would be _amazing_ if someone were to make a 3-D gif animation
of what exactly this proof is talking about, because I can't make heads or
tails of the proof language / formulae.

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theoh
It's a 2D problem: If you know the difference between circles circumscribed
around vs circles inscribed in a polygon, that's probably all the geometry you
need. The proof is a series of inequalities about the various areas inside the
circumscribed circle and outside the polygon, and also the parts of the
circumscribed circle that would be cut off by a tangent to the inscribed
circle at the points where it touches the polygon.

Does that help a bit?

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ccvannorman
Yes, in addition to the [current] top comment, this helps a great deal.
thanks!

HackerNews: Simulating AI-generated human readable proofs from real math
proofs since 2011

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jmount
The proof is short, but frankly I can't read it (as I am not sure what all the
products are sisj ; I presume they are line segments and the + is union/area
or some such). Yes, I know it because I didn't study classic geometry long
enough.

If we take the brutal step of resorting to trigonometry and calculus we can
prove the statement using general techniques. It may seem long and technical-
but I think it demonstrates re-usable methods that give some intuition (after
you use them a few times!).

To prove: the regular n-gon has the maximal area of any n-gon inscribed inside
the unit circle.

Proof sketch:

We will prove only the lemma that if theta1, and theta2 are angles summing to
C < pi then sum of the areas of the area of the figure formed by the convex
hull of (0,0), (sin(theta1),cos(theta1)), (0,1), and
(-sin(theta2),cos(theta2)) is maximized for a given C when theta1=theta2. That
is: the area formed by subdividing a segment of the circle of angle
theta1+theta2 is maximized by dividing in the middle. This would let us prove
the theorem as it implies any non-regular division can be improved by re-
centering one of the sub-divisions.

Proof of lemma:

The area of triangle formed by the convex hull of the center of a unit circle
and the points (sin(theta/2),cos(theta/2)), (-sin(theta/2),cos(theta/2)) is
(1/2) base height = (1/2) (2 sin(theta/2)) cos(theta/2) = sin(theta/2)
cos(theta/2).

So if f(theta1,theta2) is the area of the two inscribed triangles of angles
theta1 and theta2 we have: f(theta1,theta2) = sin(theta1/2) cos(theta1/2) +
sin(theta2/2) cos(theta2/2)

If we add the constraint theta1+theta2=C then we expect the maximum to occur
at one of theta1=0,theta2=C; theta1=C,theta2=0 or an point where the Lagrange
auxiliary function L(theta1,theta2,lambda) = f(theta1,theta2) + lamba
(theta1+theta2-C) has a zero gradient (gradient with respect to theta1,
theta2))

Gradient((theta1,theta2),L) = (1/2) (cos(theta1/2)^2 -sin(theta1/2)^2,
cos(theta2/2)^2 -sin(theta2/2)^2) + lambda (1,1)

For this vector quantity to be equal to zero we would have to have the
coordinates equal (so the lambda (1,1) can stomp them out): cos(theta1/2)^2
-sin(theta1/2)^2 = cos(theta2/2)^2 - sin(theta2/2)^2

But cos(x)^2 -sin(x)^2 = 1 - 2 sin(x)^2 which is a 1 to 1 function in the
range [0,pi) so to have cos(theta1/2)^2 -sin(theta1/2)^2 = cos(theta2/2)^2
-sin(theta2/2)^2 we must have theta1=theta2.

So we check that f(C/2,C/2) >= max(f(C,0),f(0,C)) and we are done.

Again I know that is long and technical (and bringing in overly big hammers)-
but the steps are all standard re-usable ideas. (actually longer than I even
intended)

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andrewla
It's worth noting that this proof implicitly uses the parallel axiom (since it
assumes a Cartesian coordinate system). The linked proof is independent of
that, and applies to non-Euclidean geometries as well.

~~~
jmount
Good point.

