number puzzle

[bobobalde111]

look dad stan said
i folded this sheet of paper so that one corner is exactly on top of the opposite corner. can you figure how long the paper is
tom looked "is it a rectangular sheet "
the boy nodded
sure the short side is 24cm and both the fold and the long side are exact number of centimetres
how long is the fold and the sheet of paper

[crofter]

Davie, This is not an easy one to PROVE, but I believe that the length of the sheet of paper is 32cms and the fold is 30cms.

These comments are based on the (3,4,5) right-angled triangle, where 3^2 + 4^2 = 5^2 (meaning 5 squared)

The sheet is then given by this triangle scaled up by a factor of 8, so that it becomes an (24,32,40) triangle; and the fold is part of this same triangle scaled up by a factor of 5 to become a (15,20,25) triangle.

The length is the 32 from the first triangle, and the fold is twice the 15 from the second triangle.

I am a bit pushed for time at the present, but I may come back to this later!

[Phil7]

Not sure how you got the answer, but it is correct...
Length = 32, Fold = 30.

The Pythagorean triple you want is 7, 24, 25, first. This give the end of the fold 7 cm. down the long edge and 25 cm. of the long edge going to the corner. So, L = 25 + 7 = 32.

Then there's a right angled triangle with hypotenuse as the fold and sides of 24 and 18 (L - 2x, or 32 - 14 = 18). This is 3,4,5 factored by 6, so length of fold = 5x6 = 30.

[crofter]

Phil7, Our methods are equivalent.

You have a sheet ABCD and fold E F (say) created by laying C onto A (so your BE = AF = 7).
You start with two Pythagorean Triangles ABE and CDF with sides (7,24,25) fitting into the Rhombus AECF.

I started with the Pythagorean Triangle CDA which is (3,4,5) factored by 8. Let AC and EF intersect at the point O. The triangle EOA is also Pythagorean (3,4,5) but factored by 5 this time.

Hence, length BC = AD = 8 x 4 = 32
and the fold is EF = 2 x OE = 2 x (5 x 3) = 30

I tried to keep my approach brief and just gave values.
I hope that things are clearer now.
What I hadn't realised was that the two triangles sitting at either end of the rhombus were also Pythagorean.
I supposed they are forced to be!

[Phil7]

Cheers, crofter, I think that while I was working on a folded diagram, you had it opened flat. So 3 stars each is pretty fair (except that you got 6 in the end!). ;-)

[crofter]

Such is life, Phil -it's not always fair. You really threw me when you began with the triple (7,24,25) when I had been working exclusively with (3,4,5).

Boba always sets a maths problem every Tuesday morning - perhaps we'll meet up again on "Boba's patch."

I'm happy to say "HONOURS EVEN"