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wu :: forums    riddles    hard (Moderators: Icarus, SMQ, ThudnBlunder, Eigenray, towr, Grimbal, william wu)    Heronian dissection: square « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Heronian dissection: square  (Read 1658 times) JocK Uberpuzzler     Gender: Posts: 877 Heronian dissection: square   « on: Aug 19th, 2005, 4:39pm » Quote Modify Can you dissect a square into as few as possible Heronian triangles?         IP Logged solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it. xy - y = x5 - y4 - y3 = 20; x>0, y>0. SMQ wu::riddles Moderator Uberpuzzler     Gender: Posts: 2084 Re: Heronian dissection: square   « Reply #1 on: Aug 20th, 2005, 6:56am » Quote Modify I see a simple solution with eight 3-4-5 right triangles of two different sizes, so that's definitely an upper bound.   --SMQ IP Logged --SMQ Oyibo Guest Re: Heronian dissection: square   « Reply #2 on: Aug 20th, 2005, 10:30am » Quote Modify Remove what the heck, for all practical purposes two triangles with edges 927538920, 927538921, 1311738121 will do.  That's definitely a lower bound. IP Logged Oyibo Guest Re: Heronian dissection: square   « Reply #3 on: Aug 20th, 2005, 10:48am » Quote Modify Remove can't find an integer solution to   (2a)2 + (a + b)2 = c2   (2a)2 + (a - b)2 = d2   (0 < b < a)   if there is, you can cut a square in three heronian triangles. IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: Heronian dissection: square   « Reply #4 on: Aug 26th, 2005, 6:14am » Quote Modify Oyibo, my sources tell that the system doesn't have non-trivial solutions.   IP Logged Sjoerd Job Postmus Full Member     Posts: 228 Re: Heronian dissection: square   « Reply #5 on: Aug 29th, 2005, 1:18am » Quote Modify I'd like to state that two is impossible, because the hypotenuse would be sqrt(2) times the length of one of the other side. sqrt(2) isn't a finite number, as far as I know.   A rectangle? Could do... A square? Nope, not in two.   So, we're looking at a minimum of three. IP Logged Oyibo Guest Re: Heronian dissection: square   « Reply #6 on: Aug 29th, 2005, 5:10am » Quote Modify Remove on Aug 29th, 2005, 1:18am, Sjoerd Job Postmus wrote:   So, we're looking at a minimum of three.   We are looking at a minimum of four (if Barukh is right).   Barukh, what makes you so sure that the zero solution is the only integer solution? IP Logged Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: Heronian dissection: square   « Reply #7 on: Aug 29th, 2005, 5:57am » Quote Modify For what it is worth, I have checked them by program.  The first "solution" it found was around a=29'000 and was due to a 32-bit integer overflow.   It is not a proof, but it convinces me. IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: Heronian dissection: square   « Reply #8 on: Aug 30th, 2005, 3:50am » Quote Modify on Aug 29th, 2005, 5:10am, Oyibo wrote: Barukh, what makes you so sure that the zero solution is the only integer solution? In your formulation, (2a, a+b, c) and (2a, a-b, d) are Pythagorean triples. We may assume then: 2a = 2mn, a+b = m2 - n2 for some integers m, n. Plugging this into the second equation, expanding and dividing both sides by n4, one gets:   x4 - 2x3 + 3x2 + 2x + 1 = y2, where x = m/n, y = d/n2 are rational numbers.   This looks threatening. Fortunately, there is a formal procedure to find rational solutions of the above equations using the method of Elliptic Curves. If I didn’t mess up with calculations, the only solution is the trivial (x, y) = (0, 1).   If there is an interest, I can elaborate on the method. There is even a software support to make things easier. IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: Heronian dissection: square   « Reply #9 on: Sep 4th, 2005, 2:53am » Quote Modify on Aug 29th, 2005, 5:10am, Oyibo wrote:   We are looking at a minimum of four (if Barukh is right). There exists a dissection into 4 parts.   I run out of time to present it now (JocK knows why  ), will make it later. IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: Heronian dissection: square   « Reply #10 on: Sep 9th, 2005, 11:23pm » Quote Modify As attachments do not work for me, I will describe a solution in words.   To get rid of fractions, let ABCD be a square with side length 360. Take point E on BC, and BC = 224; F on CD, and DF = 105. Then, all 4 triangles ABE, ADF, CEF, AFE are Heronian. IP Logged Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: Heronian dissection: square   heronian.gif « Reply #11 on: Sep 10th, 2005, 4:20am » Quote Modify Congratulations! IP Logged JocK Uberpuzzler     Gender: Posts: 877 Re: Heronian dissection: square   « Reply #12 on: Nov 1st, 2005, 9:35am » Quote Modify Excellent piece of work Barukh. Chapeau!         IP Logged solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it. xy - y = x5 - y4 - y3 = 20; x>0, y>0. Barukh Uberpuzzler     Gender: Posts: 2276 Re: Heronian dissection: square   « Reply #13 on: Nov 1st, 2005, 11:11am » Quote Modify Thanks, JocK. And welcome back!   IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium => hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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