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        riddles >> hard >> Heronian dissection: square
        
(Message started by: JocK on Aug 19^th, 2005, 4:39pm)

Title: Heronian dissection: square
Post by JocK on Aug 19^th, 2005, 4:39pm

Can you dissect a square into as few as possible Heronian triangles?

Title: Re: Heronian dissection: square
Post by SMQ on Aug 20^th, 2005, 6:56am

I see a simple solution with [hide]eight 3-4-5 right triangles of two different sizes[/hide], so that's definitely an upper bound.

--SMQ

Title: Re: Heronian dissection: square
Post by Oyibo on Aug 20^th, 2005, 10:30am

what the heck, for all practical purposes two triangles with edges 927538920, 927538921, 1311738121 will do. That's definitely a lower bound. ::)

Title: Re: Heronian dissection: square
Post by Oyibo on Aug 20^th, 2005, 10:48am

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.

Title: Re: Heronian dissection: square
Post by Barukh on Aug 26^th, 2005, 6:14am

Oyibo, my sources tell that the system doesn't have non-trivial solutions. :(

Title: Re: Heronian dissection: square
Post by Sjoerd Job Postmus on Aug 29^th, 2005, 1:18am

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.

Title: Re: Heronian dissection: square
Post by Oyibo on Aug 29^th, 2005, 5:10am

on 08/29/05 at 01:18:58, 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?

Title: Re: Heronian dissection: square
Post by Grimbal on Aug 29^th, 2005, 5:57am

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.

Title: Re: Heronian dissection: square
Post by Barukh on Aug 30^th, 2005, 3:50am

on 08/29/05 at 05:10:19, 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.

Title: Re: Heronian dissection: square
Post by Barukh on Sep 4^th, 2005, 2:53am

on 08/29/05 at 05:10:19, 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 ;D), will make it later.

Title: Re: Heronian dissection: square
Post by Barukh on Sep 9^th, 2005, 11:23pm

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.

Title: Re: Heronian dissection: square
Post by Grimbal on Sep 10^th, 2005, 4:20am

Congratulations!

Title: Re: Heronian dissection: square
Post by JocK on Nov 1^st, 2005, 9:35am

Excellent piece of work Barukh. Chapeau!

Title: Re: Heronian dissection: square
Post by Barukh on Nov 1^st, 2005, 11:11am

Thanks, JocK. And welcome back! :D