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 February 6th, 2014, 08:12 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, 08:40 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,799 Thanks: 970 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, 09:17 PM   #3
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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
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 P is gone.png (30.4 KB, 214 views)

 February 7th, 2014, 09:40 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,799 Thanks: 970 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 The "triangle(s) ABC" will retain sides 3 and 4, with the base AB changing... 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 3-4-6; point P pops up above the triangle; in the meantime, angle C reduces in size... Next integer triangle is 3-4-5: right triangle; at this time, point P coincides with point C Next is 3-4-4; now point P is inside the triangle Similarly for 3-4-3 and 3-4-2 Finally, we get degenerate 3-4-0, with AC being vertical and B on AC. Pouff! P disappears! For any scenario, the steps will similarly occur as above...
February 8th, 2014, 12:47 AM   #5
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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?
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 The wonderful life of point P.png (104.4 KB, 166 views)

February 8th, 2014, 12:51 AM   #6
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Re: A difficult problem in geometry - A Triangle and Two Squ

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 The wonderful life of point P small.png (41.2 KB, 166 views)

 February 8th, 2014, 03:52 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,104 Thanks: 1907 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, 08:59 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,799 Thanks: 970 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 Left square goes counter-clockwise, right one goes clockwise, while C rises vertically. Bye...gotta go for a P

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