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Programming Enigma Puzzles

17 July 2012

Posted by on **From New Scientist #2691, 17th January 2009** [link]

I have eight rods. I can use three of them to make the sides of a right-angled triangle of area 10 square centimetres.

Alternatively, I can use all eight of them to make the sides of an octagon, all of whose angles are the same, and all of whose vertices lie on a circle.

What is the area of that octagon?

[enigma1528]

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This one is solved without programming.

The octagon can be considered as a square with the corners cut off at 45°. So it is composed of two types of rod, of length

aandb, wherea≤b.Three of the rods can be formed into a right-angled triangle with area of 10 sq cm, so the hypotenuse has length

band the other two sides are lengtha.Hence:

a² = 20,b² = 2a² = 40, so:a= √20,b= √40.The corners cut off the octagon also form right-angled triangles with hypotenuse of length

aand the other two sides have lengthc.Hence:

a² = 2c², so:c= √10.The diagram, right, shows the octagon made up from the coloured areas, so has area:

b² + 4bc+ 2c² = 40 + 80 + 20 = 140 sq cm.Solution:The area of the octagon is 140 sq cm.Rotating the diagram through 45° gives an even more straightforward depiction of the solution, and directly includes the

aabright-angled triangle of area 10 sq cm in the corners (and also as the removed corners).Giving the area as: 5

a² + 4a²/2 = 7a² = 140 sq cm.Or considering the removed corners: 9

a² – 4a²/2 = 7a² = 140 sq cm.