wu :: forums
« wu :: forums - Dijkstra, AMM and geometry »

Welcome, Guest. Please Login or Register.
Apr 25th, 2024, 4:21pm

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   medium
(Moderators: Eigenray, Icarus, Grimbal, SMQ, william wu, ThudnBlunder, towr)
   Dijkstra, AMM and geometry
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: Dijkstra, AMM and geometry  (Read 6987 times)
Aryabhatta
Uberpuzzler
*****






   


Gender: male
Posts: 1321
Dijkstra, AMM and geometry   dijkstra_amm.PNG
« on: Apr 19th, 2007, 10:58am »
Quote Quote Modify Modify

A geometry problem which interested Dijkstra, from AMM 1998.
 
(AMM = American Mathematics Monthly)
 
 
In the figure below,
 
If AD + DF equals AB + BF
 
Show that
 
AE + EF = AC + CF
 
« Last Edit: Apr 19th, 2007, 10:58am by Aryabhatta » IP Logged

rmsgrey
Uberpuzzler
*****





134688278 134688278   rmsgrey   rmsgrey


Gender: male
Posts: 2873
Re: Dijkstra, AMM and geometry  
« Reply #1 on: Apr 20th, 2007, 8:26am »
Quote Quote Modify Modify

One observation which may lead nowhere: A and F are focii of an ellipse, and B and D points on the perimeter.
IP Logged
Barukh
Uberpuzzler
*****






   


Gender: male
Posts: 2276
Re: Dijkstra, AMM and geometry   Urquahart1.PNG
« Reply #2 on: Apr 27th, 2007, 9:51am »
Quote Quote Modify Modify

Here’s a nice solution I learned elsewhere.
 
If we can show that, under given conditions, there exists a circle touching four lines AC, AD, CD, BE (prolonged, if needed – see the drawing), that the statement of the problem will follow (why?)
 
To show that such a circle indeed exists, consider the ex-circles of triangles ACD and ABE (the ex-circle of a triangle is a circle tangent to 2 of its sides internally and one externally).  
 
The center O1 of the ex-circle of ACD is the intersection of angle bisectors A and CDE.
The center O2 of the ex-circle of ABE is the intersection of angle bisectors A and CBE.
 
Now, consider the triangle FF1F2, where F1, F2 are chosen on corresponding lines such that BF1 = BF, DF2 = DF. Since AF1 = AB+BF1 = AB+BF = AD+DF = AD+DF2 = AF2, we get that all three triangles F1AF2, F1BF, F2DF are isosceles.
 
Therefore, the bisectors of angles A, CBE, CDE coincide with the bisectors of corresponding segments F1F2, FF1, FF2. But the latter 3 bisectors are concurrent, since they define the center of the circumcircle of the triangle F1FF2. Thus, O1 and O1 coincide.
 
Finally, the ex-circles of triangles ACD and ABE also coincide, since they touch the same two lines and are centered at the same point. But now it follows that this circle touches all four needed lines!
« Last Edit: Apr 27th, 2007, 10:03am by Barukh » IP Logged

Aryabhatta
Uberpuzzler
*****






   


Gender: male
Posts: 1321
Re: Dijkstra, AMM and geometry  
« Reply #3 on: Apr 27th, 2007, 12:49pm »
Quote Quote Modify Modify

Interesting solution Barukh!
IP Logged
Barukh
Uberpuzzler
*****






   


Gender: male
Posts: 2276
Re: Dijkstra, AMM and geometry  
« Reply #4 on: Apr 27th, 2007, 11:10pm »
Quote Quote Modify Modify

Thanks. Some interesting results about this theorem:
 
1. It is attributed to Australian mathematician M. L. Urguhart who discovered the result in 1960s while "considering some of the fundamental concepts of the theory of special relativity" (!)
 
2. The result was proved as early as 1841 by English mathematician De Morgan.
 
3.  The theorem has the following nice equivalent statement (partly spotted by rsmgrey): If C and E are points on an ellipse with foci A and F, then points of intersection AC, EF and AE, CF lie on a confocal ellipse".
IP Logged
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board