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        riddles >> easy >> Integral Square
        
(Message started by: Sir Col on Aug 29^th, 2007, 9:55am)

Title: Integral Square
Post by Sir Col on Aug 29^th, 2007, 9:55am

A right triangle has legs with integer lengths x and y.
A square is placed inside the triangle with one vertex at the right angle and the opposite vertex on the hypotenuse.
Determine the conditions for the square to have an integral side length.

Title: Re: Integral Square
Post by ThudanBlunder on Aug 29^th, 2007, 10:08am

[hide]x+y divides xy[/hide]

Title: Re: Integral Square
Post by ecoist on Aug 31^st, 2007, 3:24pm

[hide]The existence alone of a square so located inside a right triangle restricts x and y.[/hide]

Title: Re: Integral Square
Post by Sameer on Aug 31^st, 2007, 4:56pm

I get [hide] If a is length of square, then a² = (x-a)*(y-a) [/hide]. I can't think of what to express it in words yet...

Title: Re: Integral Square
Post by ecoist on Sep 1^st, 2007, 6:27pm

What integers satisfy ThundanBlunder's condition? If a and b are arbitrary positive integers, then x=a(a+b) and y=b(a+b) satisfies his condition, with a side of the inscribed square of length ab. But, alas, x=18 and y=9 also satisfies ThudanBlunder's condition, with square of side length 6, yet the above formula doesn't cover this case.

Title: Re: Integral Square
Post by Eigenray on Sep 1^st, 2007, 7:13pm

An integer parametrization is [hide]x = ka(a+b), y = kb(a+b), where gcd(a,b)=1[/hide].

Title: Re: Integral Square
Post by Sameer on Sep 1^st, 2007, 7:21pm

on 09/01/07 at 18:27:31, ecoist wrote:

yet the above formula doesn't cover this case.

I see. It's weird. I started with this:
Square of length a
x = a+b
y = a+c
So sqrt(a² + c²) + sqrt(a² + b²) = sqrt( (a+c)² + (a+b)²)

Simplifying which gives me a² = bc

I wonder why doesn't the case of x=18 and y = 9 get covered!! :-/

Title: Re: Integral Square
Post by Grimbal on Sep 2^nd, 2007, 2:53pm

on 09/01/07 at 19:21:27, Sameer wrote:

I wonder why doesn't the case of x=18 and y = 9 get covered!! :-/

Doesn't it?
a = 6, b = 3, c = 12
a² = b·c = 36
x = a+b = 9, y = a+c = 18

Title: Re: Integral Square
Post by Sameer on Sep 2^nd, 2007, 3:00pm

on 09/02/07 at 14:53:37, Grimbal wrote:

Isn't it?
a = 6, b = 3, c = 12
a² = b·c = 36
x = a+b = 9, y = a+c = 18

Duh!?! Thanks!! Seems like I have been overworking again!! However my original question still holds. How do you geometrically or in words express a² = bc!!

Title: Re: Integral Square
Post by Grimbal on Sep 2^nd, 2007, 3:01pm

Inserting the square creates 2 smaller triangles, one with sides b and a, the other with sides a and c. The fact that they are similar implies b/a = a/c, which directly gives a² = bc

Title: Re: Integral Square
Post by Sir Col on Sep 2^nd, 2007, 3:07pm

And using similarity, (x-a)/a = a/(y-b), which gives a² = (x-a)(y-a) = xy-a(x+y)+a², leading to T&B's result: a = xy/(x+y).

Title: Re: Integral Square
Post by Grimbal on Sep 2^nd, 2007, 3:09pm

The fact that the 2 smaller trianges are similar implies
(x-a)/a = a/(y-a)
<=> (x-a)(y-a) = a²
<=> xy-ax-ay-a² = a²
<=> xy-a(x+y) = 0
<=> a = xy/(x+y)
So, for a to be an integer, xy must be a multiple of (x+y), which conveniently removes the unknown a from the equation.
And it proves what ThudanBlunder said straight away and Sir Col minutes ago. :-/

Title: Re: Integral Square
Post by Sameer on Sep 2^nd, 2007, 3:11pm

Thanks for the great simple explanation!! :)I unnecessarily seem to make things complicated, don't I? :-[

Title: Re: Integral Square
Post by Sir Col on Sep 2^nd, 2007, 3:12pm

on 09/02/07 at 15:09:51, Grimbal wrote:

(sigh!)

At least you proved it. I think that T&B's was just a lucky guess. ;)

Title: Re: Integral Square
Post by ThudanBlunder on Sep 3^rd, 2007, 7:59pm

on 09/02/07 at 15:12:12, Sir Col wrote:

At least you proved it. I think that T&B's was just a lucky guess. ;)

And the great thing is the more I study, the luckier I get. :P