**From New Scientist #2798, 5th February 2011** [link]

I started with a rectangular piece of paper, the longer side of which was 24 centimetres in length. I then folded the paper with one straight fold so that one corner of the rectangle was touching the diagonally opposite corner. This created a folded piece that was a pentagon, and the area of this pentagon was two-thirds of the area of the original rectangle.

What was the length of the longest side of the pentagon?

[enigma1632]

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This one doesn’t really need any programming. The solution falls out with a little analysis…

Consider the unfolded rectangle. It has an area of 24

a(whereais the length of the shorter side),and is composed of 4×

T1triangles (on the outsides) and 2×T2triangles (in the middle).Then consider the folded pentagon. It has an area 2/3 of the rectangle, so

16a, and iscomposed of 2×

T1triangles and aT1+T2triangle.So considering the areas:

4 × T1 + 2 × T2 = 24a3 × T1 + T2 = 16ahence:

T1andT2both have area4a.And they both have height

a, so they are identical triangles (T) with a base measuring 8 (=b).Which means the longest side of the pentagon is the hypotenuse of

T(=h), the other sides of the pentagon being the non-hypotenuse sides ofT(aandb).But when we make the fold the bases of two

Ttriangles lie along the hypotenuse of anotherTtriangle it follows thath = 2 × b, henceh = 16(anda = √192).Solution:The longest side of the pentagon is 16cm.