The Grey Cells - Exercise !

Originally posted by gcle2003 I got led astray a while by the coincidence that a 3,4,5 right-angled triangle has perimeter 12 though sadly area 6.

True, one can take the right triangle with sides 3, 4 and 5 and then 'move' the match sticks so as to 'cut out' 2 squares.  For example, assume that the base is the 3 side, the perpendicular side is 4 and the hypotenuse is of course 5.  Then take the 'bottom' 2 sticks of the 4 side, go one match stick towards the hypotenuse then down 2 to meet the bottom.  However, to 'prove' that is a proper polygon, one needs to 'prove' geometrically that the 'indentation' does not in fact reach to or past the hypotenuse.  To be honest, I do not find that to be a more 'obvious' solution than the parallelogram.  However, both are solutions to the problem as originally stated.

I agree one can 'see it'.  However, in the theoretical world of pure mathematics, we cannot accept visual evidence as 'proof'.  I can 'see' the result before I 'prove' it, however, it is a non-trivial matter to 'prove' that the line (2 match sticks) that 'cuts into' the triangle does not in fact touch the hypotenuse before it 'turns' down to the base.  That is of course a necessary condition to 'prove' that the shape is in fact a single polygon.  Personally I 'see' any solution involving match stick not being either parallel or perpendicular as being 'topologically' equivalent to a 'triangular' solution.  Of the 'triangular' solutions, I prefer the parallelogram.  Interestingly, the parallelogram described (with sides 2 and 4) can be manipulated to form any area between 8 (which would be a regular rectangle ) to zero (completely collapsed to a line).  I will try to post a picture once I figure out how to convert what I've done into one of the perscribed formats.  

Originally posted by Northman   Bingo - thats it. You don't have to prove anything - you can see it directly:     you're up again deadkenny

Very clever solution.  My thought was that you could form a smaller triangle by extending your line DE to AB and then use trig to show that that side of the triangle is longer than 1, which is the known length of DE.  However your 'line equation' method simply using the assigned coordinates of the vertices is much 'cleaner'.  

Edited by deadkenny - 06-Aug-2008 at 16:07

It occurs to me I maybe should clarify that it is permissible - indeed necessary - to attach more than two matches at the same point, so that you can, e.g., make a 'Y' as well as a 'V', as long as it is always at the ends. 

I am implicitly assuming that we are working in 2 dimensions here.  My first take is a rather 'ugly' solution.  We start with the old 3x4x5 right triangle.  Now of course all of those 'flexible joints' allow it to be 'deformed'.  So we 'butress' the sides with interlocking equalateral triangles. If I counted correctly the total match stick usage is 45.  Now while I believe that satisfies the 'rigid right angle' requirement, I doubt that it does so with the minimium number of matchsticks necessary.

That's the approved answer. Of course you can get a rigid 90° angle using any other Pythagorean triangle, but they all use more matches essentially because they all have greater perimeters (the number of matches for perimeter p is 4p-3).