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Great god.... even esoteric subject get a fine treatment. Thanks to whoever contributed to this article. Xah Lee

adding dynamic geometry program files

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btw, there's this great free software:

Compass and Ruler is written in Java by von R Grothmann. This program is excellent. It is very well written, in Java and available for Mac or Windows or Linux, and it is free with source code too. Its home page is at http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/index.html

that we can use and build example files to go along with article and theorems. Xah Lee 15:36, 2005 Mar 18 (UTC)


I don't understand from the proof... I can't see the equivalence with those scalar & cross products.


I also have trouble with the proof. It depends on various lines being "concurrent," but "concurrent is nowhere defined. — Preceding unsigned comment added by 119.252.188.83 (talk) 13:08, 19 August 2015 (UTC)[reply]

application in perspective drawing

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There is a nice practical application in perspective drawing: suppose you have two (or more) lines drawn in perspective, which meet at a vanishing point located far enough away from the physical page to be constructed directly. Desargues' theorem can be used to construct additional lines which meet at the same vanishing point. --Heath 69.174.67.197 05:04, 16 July 2006 (UTC)[reply]

Suggested merger from Perspective (geometry)

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I needed a link for perspective triangles recently and was forced to resort to MathWorld instead, as the Perspective (geometry) article hadn't yet been created. I would prefer not to merge, as I think that perspective triangles are important enough, that there are enough results about them that are distinct from Desargues' theorem, and that there are enough other topics already included in the Desargues theorem article that it would be confusing to link to it under the perspective triangle name. —David Eppstein (talk) 04:55, 28 December 2007 (UTC)[reply]

Specifically, what results about perspective triangles other than Desargues' theorem do you have in mind? Michael Hardy (talk) 22:11, 5 January 2008 (UTC)[reply]
A better question might be: Specifically what other relations of Desargues' theorem to concepts in mathematics do you have in mind other than perspective triangles? The theorem should be merged into the definition if anything. Nousernameslefttalk and matrix? 02:45, 13 January 2008 (UTC)[reply]
Since this hasn't any opposition for quite a while, I'll just remove the template. Nousernameslefttalk and matrix? 18:44, 5 January 2008 (UTC)[reply]

Desargues' or Desargues's?

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Not a matter for concern on a par with the current state of the economy, but Wikipedia's style guide says:

Proper nouns that end with s: There is tension in English over whether just an apostrophe, or an apostrophe and the letter s, should be added to such proper nouns (James' house or James's house, but be consistent within an article). Some forms always take an extra s (Ross's father); some, mainly Biblical and Classical, usually do not (Socrates' wife; Moses' ascent of Sinai; Jesus' last words).

Although omitting the s used to be the norm for such names (I was raised in the pro-omission tradition and have been accustomed until recently to writing it as Desargues'), more recent sources such as the 1961 edition of Coxeter's Introduction to Geometry, The Concise Oxford Dictionary of Mathematics, and Britannica (at least as far back as 1986, the publication year of my hard copy) write it as "Desargues's theorem," as do the more authoritative sources on the web, although a majority of pop writers still seem to be using the old style. Journal editors these days seem to be encouraging its inclusion (I was recently picked on by one for using the old style and adapted accordingly).

Does anyone have any strong opinions either way? In particular are there any strong objections to matching Wikipedia's style in the case of Desargues to the usage of the sources cited above? --Vaughan Pratt (talk) 02:08, 8 December 2008 (UTC)[reply]

Non-Desargues configuration

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It is possible for ten points and ten lines in the plane to be so arranged that each of the ten lines passes through three of the ten points and each of the ten points lies on three of the ten lines, even though that whole configuration is not incidence-isomorphic to the Desargues configuration. Should that be mentioned here?

Michael Hardy (talk) 18:22, 23 August 2009 (UTC)[reply]

Added. As always, feel free to reword my addition. —David Eppstein (talk) 22:24, 23 August 2009 (UTC)[reply]

Extended Desargues' theorem and Its Application in Computer Graphics

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§ First, let Desargues theorem be simplified as the symbol of , where and , and , and are corresponding point pairs respectively, is the intersection point of the lines , and .

§ Then the extended Desargues' theorem in higher dimensional projective space, e.g., three dimensional projective space, can be represented as . and are tetrahedrons or at most one of them is quadrangle. In this case, the intersection points of line pairs and , , and , are coplanar, not collinear any more.

§ Detailed proof via homogeneous coordinates representation of the extended Desargues theorem, and its applications in computer graphics can be seen from the link below:

http://www.newsmth.net/att.php?p.50.48793.475.pdf

§ A series of geometric transformations, perspective projection, parallel projection, reflection, translation transform and so on, can all be defined as one concept: , of which transform matrix has been proved to be equivalent to the concept of in numeric analysis.

So these geometric transforms can be easily be written in elementary matrices coordinate system independently, i.e., we don't have to project objects only to those coordinate planes in order to simplify the projection matrix establishment procedure: the projection matrix which projects objects to an arbitrary plane is actually an elementary matrix in very simple form which we can easily write it out!

§ The applications of the above results were further discussed in the link mentioned.

Cite error: There are <ref> tags on this page without content in them (see the help page).http://www.newsmth.net/att.php?p.50.48793.475.pdf

File:Desargues theorem alt.svg to appear as POTD soon

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Hello! This is a note to let the editors of this article know that File:Desargues theorem alt.svg will be appearing as picture of the day on November 2, 2011. You can view and edit the POTD blurb at Template:POTD/2011-11-02. If this article needs any attention or maintenance, it would be preferable if that could be done before its appearance on the Main Page so Wikipedia doesn't look bad. :) Thanks! howcheng {chat} 20:34, 31 October 2011 (UTC)[reply]

Desargues' theorem
In projective geometry, Desargues' theorem states that two triangles are in perspective axially if and only if they are in perspective centrally. Lines through the triangle sides meet in pairs at collinear points along the axis of perspectivity. Lines through corresponding pairs of vertices on the triangles meet at a point called the center of perspectivity.Image: DynaBlast

I have two concerns, although I'm not sure of the best way to address them.

First, in the picture caption, I think the phrase when produced is likely to be confusing to non-mathematicians.

Second, the lead paragraph beginning Desargues's theorem holds for ... is somewhat clumsy. I'm used to thinking about Desargues's theorem as a result about projective planes: two triangles can't be in axial perspective unless there exists a plane containing both. So if we replace "projective spaces" by "projective planes" throughout the article then there are fewer cases to explain in this lead paragraph. But that would be a significant rewrite, and I'm not sure whether it's justified by the sources. Jowa fan (talk) 01:13, 1 November 2011 (UTC)[reply]

I agree that the lead could use a rewrite, but I can't do it in the amount of time left. The suggestion about replacing spaces by planes doesn't work, since it is precisely that two triangles in different planes which are centrally perspective are automatically axially perspective which makes the proof in 3 dimensions so easy. I've done what I can for this article - if only I had an additional 24 hours or had this occurred in a different week - Oh well! Bill Cherowitzo (talk) 16:04, 1 November 2011 (UTC)[reply]

I edited the POTD caption to avoid using the word "produced". I agree that it is exactly the right technical term for the line that contains a line segment, but we should avoid jargon that is likely to be unfamiliar to non-geometers. —David Eppstein (talk) 17:35, 1 November 2011 (UTC)[reply]

Maybe you thought of a solution but didn't actually implement it. The caption still said "produced" when I looked at it. I've made a change, but feel free to edit this further if you can think of a better way. Jowa fan (talk) 23:27, 1 November 2011 (UTC)[reply]
Sorry, I meant that I changed it in the POTD box. I didn't make the same change to the actual article. —David Eppstein (talk) 23:56, 1 November 2011 (UTC)[reply]

This article needs to include dates

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Please add the date when this theorem was discovered or proved or conjectured --Thepurplelefant (talk) 12:55, 8 June 2012 (UTC)[reply]

Done Bill Cherowitzo (talk) 01:07, 9 June 2012 (UTC)[reply]

Permission to use image

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I'd like to use the image on the t-shirts we give away to students in out local high school mathematics contest. I don't really want to write an attribution on the shirt. Would that be okay? — Preceding unsigned comment added by 142.27.68.70 (talk) 19:20, 31 March 2015 (UTC)[reply]

The CC license for this image does call for attribution when it is used (remixed in your case). Try Googling Desargues theorem images, I'm sure that you'll find one that doesn't ask for attribution. Bill Cherowitzo (talk) 22:08, 31 March 2015 (UTC)[reply]
You might also try looking at some of the other images in Commons:Category:Desargues' theorem. For instance, File:DesarguesHessenberg.svg has a license that doesn't require attribution, and might fit your needs. —David Eppstein (talk) 22:11, 31 March 2015 (UTC)[reply]

Thanks to both of you for your advice.Aliotra (talk) 12:53, 2 April 2015 (UTC)[reply]

Definition of concurrent

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The 3-dimesional proof depends on various lines being "concurrent." Please define "concurrent." — Preceding unsigned comment added by 119.252.188.83 (talk) 13:02, 19 August 2015 (UTC)[reply]

 Done Bill Cherowitzo (talk) 15:48, 19 August 2015 (UTC)[reply]

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Proof of converse statement?

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The article hand-waves it away by saying it follows by duality. But some projective planes are not self-dual. --Svennik (talk) 00:41, 7 February 2024 (UTC)[reply]

It doesn't matter whether the projective plane is self-dual; what matters is that the statement is self-dual. That is, the dual of the statement is the same statement. So if you prove the statement in some plane (self-dual or not) you have also proved the dual of the statement. —David Eppstein (talk) 01:14, 7 February 2024 (UTC)[reply]