Author 
Topic: Dijkstra, AMM and geometry (Read 5276 times) 

Aryabhatta
Uberpuzzler
Gender:
Posts: 1327

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 19^{th}, 2007, 10:58am by Aryabhatta » 
IP Logged 



rmsgrey
Uberpuzzler
Gender:
Posts: 2701


Re: Dijkstra, AMM and geometry
« Reply #1 on: Apr 20^{th}, 2007, 8:26am » 
Quote 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:
Posts: 2273

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 excircles of triangles ACD and ABE (the excircle of a triangle is a circle tangent to 2 of its sides internally and one externally). The center O_{1} of the excircle of ACD is the intersection of angle bisectors A and CDE. The center O_{2} of the excircle of ABE is the intersection of angle bisectors A and CBE. Now, consider the triangle FF_{1}F_{2}, where F_{1}, F_{2} are chosen on corresponding lines such that BF_{1} = BF, DF_{2} = DF. Since AF_{1} = AB+BF_{1} = AB+BF = AD+DF = AD+DF_{2} = AF_{2}, we get that all three triangles F_{1}AF_{2}, F_{1}BF, F_{2}DF are isosceles. Therefore, the bisectors of angles A, CBE, CDE coincide with the bisectors of corresponding segments F_{1}F_{2}, FF_{1}, FF_{2}. But the latter 3 bisectors are concurrent, since they define the center of the circumcircle of the triangle F_{1}FF_{2}. Thus, O_{1} and O_{1} coincide. Finally, the excircles 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 27^{th}, 2007, 10:03am by Barukh » 
IP Logged 



Aryabhatta
Uberpuzzler
Gender:
Posts: 1327


Re: Dijkstra, AMM and geometry
« Reply #3 on: Apr 27^{th}, 2007, 12:49pm » 
Quote Modify

Interesting solution Barukh!


IP Logged 



Barukh
Uberpuzzler
Gender:
Posts: 2273


Re: Dijkstra, AMM and geometry
« Reply #4 on: Apr 27^{th}, 2007, 11:10pm » 
Quote 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 



