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Programming Enigma Puzzles

10 September 2013

Posted by on **From New Scientist #2553, 27th May 2006**

I designed a water feature for my garden, fenced in for safety. I planned to buy three pieces of fencing to form a triangular boundary, with two of them being equal in length. Then I planned to make the biggest possible circular pool within the triangle, seeding the rest of the area with grass.

In my haste at the garden centre I bought the two correct equal-length pieces, but the third piece was only three-quarters as long as it should have been. I constructed a triangle with the incorrect fencing and made the largest circular pond possible within it.

It turned out that the area enclosed by the triangle was exactly the same as I had intended in my original plan but, surprisingly, the radius of the pool I ended up with was one foot greater than I had originally planned. What three lengths of fencing did I actually buy?

[enigma1393]

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Consider the planned triangle (T

_{0}) to have sides 8a, 8a, 8b. The semi-perimeter is 8a + 4b, and hence the area A, by Heron’s formula, is:The actually constructed triangle (T

_{1}) has sides 8a, 8a, 6b. The semi-perimeter 8a + 3b, and the area is also A. Heron’s formula gives:Hence:

which simplifies to:

So the semi-perimeter of T

_{0}is 9b, and the semi-perimeter of T_{1}is 8b.Now, considering the incircles of the triangles. The area of a triangle is the product of the inradius and the semi-perimeter, so:

We see that the triangles are arranged as below:

And so the area is:

And so the sides of T

_{0}are (30, 30, 48) and the sides of T_{1}are (30, 30, 36), which is the required answer.Solution:The lengths of fencing bought were two lengths of 30 feet and one length of 36 feet.