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February 6th, 2014, 07:12 AM  #1 
Newbie Joined: Aug 2013 Posts: 19 Thanks: 0  A difficult problem in geometry  A Triangle and Two Squares http://agutie.homestead.com/files/tr..._square_01.htm Given that everything that had to be demonstrated in the problem above was indeed demonstrated, solve this: M is the midpoint of segment EF. Prove that the quadrilateral AMPC (or APMC, depends on how you drew triangle ABC) is a cyclic quadrilateral (or that points A,M,P,C lie on the same circle). Thank you. 
February 6th, 2014, 07:40 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,106 Thanks: 796  Re: A difficult problem in geometry  A Triangle and Two Squ
I quit...ran out of Tylenols My diagram has more lines than a sheet of graph paper :P Btw, you do realise that point P can end up outside triangle ABC, right? 
February 6th, 2014, 08:17 PM  #3 
Newbie Joined: Aug 2013 Posts: 19 Thanks: 0  Re: A difficult problem in geometry  A Triangle and Two Squ
HA! You're right! And not only outside the triangle ABC, but it may not exist at all! (if CD and AG are not concurrent as demonstrated below) Maybe with another configuration, the answer will somehow show itself 
February 7th, 2014, 08:40 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,106 Thanks: 796  Re: A difficult problem in geometry  A Triangle and Two Squ
Son of a gun, this is big fun... We can start like this (the 2 squares only; using 3by3 and 4by4 to illustrate): Code: G 4 F 1 C E 3 D 4 3 4 3 3 B 3 C 4 A B ? A Point P is invisible at start, ready for action! No triangle yet... Point C rises vertically, forcing AB to shorten from 7: the 1st possible integer triangle is 346; point P pops up above the triangle; in the meantime, angle C reduces in size... Next integer triangle is 345: right triangle; at this time, point P coincides with point C Next is 344; now point P is inside the triangle Similarly for 343 and 342 Finally, we get degenerate 340, with AC being vertical and B on AC. Pouff! P disappears! For any scenario, the steps will similarly occur as above... 
February 7th, 2014, 11:47 PM  #5 
Newbie Joined: Aug 2013 Posts: 19 Thanks: 0  Re: A difficult problem in geometry  A Triangle and Two Squ
I got that P is still there for the degenerate case. Did I misunderstand your construction?

February 7th, 2014, 11:51 PM  #6 
Newbie Joined: Aug 2013 Posts: 19 Thanks: 0  Re: A difficult problem in geometry  A Triangle and Two Squ
Too big, sorry about that 
February 8th, 2014, 02:52 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,844 Thanks: 1566 
In the first diagram, EC can be extended to meet BG and DF at P. In each case, EC and BG are perpendicular and DF is at 45° to each of them. Also, BG = EC = DF/?2.

February 8th, 2014, 07:59 AM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,106 Thanks: 796  Re: A difficult problem in geometry  A Triangle and Two Squ
Nice diagram! We're really doing same thing, except my initial ABC is your last ABC; I rotate both squares...but not having your diagramatic skills, can't clearly show what I'm doing. Code: B 3 C 4 A : start C B 6 A : 1st integer ABC while C rises vertically. Bye...gotta go for a P 

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