Mathematics B‑class Low‑priority

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Religious belief?

Section. —Nils von Barth (nbarth) (talk) 21:59, 20 February 2010 (UTC)[reply]

Where is the mention of Pascal's theory of the logic of religious belief? That's one of his most widely quoted theorems. — Preceding unsigned comment added by 71.146.44.135 (talk)

That's not a theorem. Maybe you should start by looking that the article titled Blaise Pascal. Michael Hardy (talk) 23:27, 18 November 2007 (UTC)[reply]

Sixth Step

What happened to the sixth step?

http://en.wikipedia.org/w/index.php?title=Pascal%27s_theorem&diff=prev&oldid=192916740

--Scottdavies (talk) 13:49, 27 September 2008 (UTC)[reply]

And the second step?

Complicated proof

Why is such a messy proof being used in this article? It looks a very short proof is being linked using Menelaos. A similarly short proof can also be found as Theorem 6.3.1 here: http://www-math.mit.edu/~kedlaya/geometryunbound/gu-060118.pdf. 76.69.85.111 (talk) 16:18, 18 December 2008 (UTC)[reply]

I added a short proof using projective geometry which I think is instructive as it applies to the original conic as well. Lim Wei Quan (talk) 14:59, 20 November 2009 (UTC)[reply]

Unnecessarily complex picture

The diagram illustrating this article is unnecessarily complicated, because the hexagon has been chosen to be 'tangled-up'.

Although the theorem is of course true in this case too, this will make most readers miss the point.

A simpler and better picture is http://commons.wikimedia.org/wiki/File:THPascal.svg

84.97.149.75 (talk) 11:28, 13 February 2009 (UTC)[reply]

Much clearer – thanks! I’ve added it.

—Nils von Barth (nbarth) (talk) 22:04, 20 February 2010 (UTC)[reply]

polar dual

Pascal's theorem is the polar dual, not projective dual, of Brianchon's theorem, it seems to me. Does anyone have a source for this? Tkuvho (talk) 18:03, 9 March 2010 (UTC)[reply]