[Search for users] [Overall Top Noters] [List of all Conferences] [Download this site]

Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

2038.0. "Whose theorem is this?" by IOSG::MAURICE (Like a tea tray in the sky) Wed Apr 10 1996 11:35

    Hi,
    
    Does anyone know whose theorem this is:
    
    You have two straight lines. On the first line there are points A, B
    and C. On the second line there are points L, M & N. The points of
    intersection of where (AM & BL), (AN & CL), & (BN & CM) meet, turn out
    to be on a straight line themselves.
    
    Thanks
    
    Stuart
T.RTitleUserPersonal
Name		DateLines
2038.1HANNAH::OSMANsee HANNAH::IGLOO$:[OSMAN]ERIC.VT240Wed Apr 10 1996 14:287
Does this have something to do with the pantograph, a drawing instrument looking
kind of like a tresseled wooden stairwell gate lieing down horizontally,
that allows a user to trace a drawn figure and produce a larger or smaller
replica ?

/Eric
2038.2Desargues?FLOYD::YODERMFYWed Apr 10 1996 21:402
The name "Desargues" comes to mind, but I may be
confusing this with another theorem.
2038.3IOSG::MAURICELike a tea tray in the skyThu Apr 11 1996 07:4416
    Hi,
    
    Thanks for the hints. I tried searching the net using them , and it
    does seem that the theorem of Desargues is close. It says "If two
    triangles are in perspective, then the meets of the corresponding sides
    are collinear". From:
    
    http://www-groups.dcs.st-and.ac.uk/~history/BigPictures/DesarguesTheorem.gif
    
    (Which also has a picture)
    
    My maths isn't good enough to know whether this is the same theorem.
    
    Cheers
    
    Stuart
2038.4PappusIOSG::MAURICELike a tea tray in the skyThu Apr 11 1996 08:138
    A lucky search on the net using the Alta Vista search engine for
    +collinear +theorem gave up Pappus of Alexandria. " If points A1, A2,
    A3 and B1, B2, B3 are collinear, then the intersections A1B2.A2B1,
    A1B3.A3B1, A2B3.A3B2 are collinear."
    
    Cheers
    
    Stuart
2038.5HANNAH::OSMANsee HANNAH::IGLOO$:[OSMAN]ERIC.VT240Thu Apr 11 1996 19:265
And where is the newly defined line.  Does it bisect the angle formed
by the original two ?

/Eric
2038.6hand wavingAUSS::GARSONachtentachtig kacheltjesThu Apr 11 1996 22:3716
    re .5                                
    
>Does it bisect the angle formed by the original two ?
    
    I wouldn't think so.
    
    Let one line be the x-axis with points at x=1,2,3 (call them A,B,C) and
    the other line be the y-axis with points at y=1,2,3 (call them D,E,F).
    The point formed from A,B,D,E (call it G) indeed is on the bisector, as
    are the other two points formed. Now move the point F up the y-axis.
    The point G does not move because it is not determined by F (and hence
    remains on the bisector). However the other two points (call them H,I)
    clearly do move. It is intuitively obvious (proof by declaration) that
    H (and I) both move off the bisector (because only one of the lines
    determining their position is moving) and hence G,H,I is no longer the
    bisector.