4 ms·

New proof of a minimum property of the regular n-gon (1947)

41 points by mgdo 11y ago

ccvannorman 11y ago
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.
- nilkn 11y ago
  I recommend just reading the attached notes and imagery rather than the document itself. They've added pictures that show what it's talking about and clarified parts of the argument that were left vague. I think they do a much better job of explaining the proof than the paper does. This is a good example of a proof that was written to be as short as possible rather than as clear as possible.
- theoh 11y ago
  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?
  - ccvannorman 11y ago
    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
- j2kun 11y ago
  You have to read it actively. Draw your own pictures :)
deleted 11y ago
[deleted]
GFK_of_xmaspast 11y ago
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.
- kens 11y ago
  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 11y ago
    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)
  - jessriedel 11y ago
    Tangentially, I found this notable for the reasonably high-equality, in-browser PDF commenting.
- dang 11y ago
  Thanks—added.
jmount 11y ago
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)
- nilkn 11y ago
  > I am not sure what all the products are sisj These symbols are intended to refer to the areas of overlap between the circular sections s_i and s_j. It is an odd notation and it's not one that I've seen before. You can infer what it means though when earlier in the paper they refer to the "product" C * P_n as the "part of P_n lying in C".
cousin_it 11y ago
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.
- SilasX 11y ago
  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.

41 points by mgdo 11y ago

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.

nilkn 11y ago
I recommend just reading the attached notes and imagery rather than the document itself. They've added pictures that show what it's talking about and clarified parts of the argument that were left vague. I think they do a much better job of explaining the proof than the paper does. This is a good example of a proof that was written to be as short as possible rather than as clear as possible.
theoh 11y ago
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?
- ccvannorman 11y ago
  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
j2kun 11y ago
You have to read it actively. Draw your own pictures :)

deleted 11y ago

[deleted]

GFK_of_xmaspast 11y ago

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.

kens 11y ago
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 11y ago
  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)
- jessriedel 11y ago
  Tangentially, I found this notable for the reasonably high-equality, in-browser PDF commenting.
dang 11y ago
Thanks—added.

jmount 11y ago

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)

nilkn 11y ago
> I am not sure what all the products are sisj These symbols are intended to refer to the areas of overlap between the circular sections s_i and s_j. It is an odd notation and it's not one that I've seen before. You can infer what it means though when earlier in the paper they refer to the "product" C * P_n as the "part of P_n lying in C".

cousin_it 11y ago

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.

SilasX 11y ago
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.