# Enigmatic Code

Programming Enigma Puzzles

## Enigma 1632: Foiled by a fold

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]

### One response to “Enigma 1632: Foiled by a fold”

1. jimrandell 13 December 2011 at 1:43 pm

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 24a (where a is the length of the shorter side),
and is composed of 4× T1 triangles (on the outsides) and 2× T2 triangles (in the middle).

Then consider the folded pentagon. It has an area 2/3 of the rectangle, so 16a, and is
composed of 2× T1 triangles and a T1+T2 triangle.

So considering the areas:
4 × T1 + 2 × T2 = 24a
3 × T1 + T2 = 16a
hence: T1 and T2 both have area 4a.

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 of T (a and b).

But when we make the fold the bases of two T triangles lie along the hypotenuse of another T triangle it follows that h = 2 × b, hence h = 16 (and a = √192).

Solution: The longest side of the pentagon is 16cm.

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