## Teaser 1973: Straw store

**From The Sunday Times, 9th July 2000** [link]

Farmer Fermat has more straw than he can safely stack on his usual rectangular storage area. So he extends the shorter sides of the area to equal the longer sides, thereby forming a square. Alongside he adds another square area, whose side is equal to the shorter side of the original rectangle. The total area of the two squares together is 243 square feet larger than the original rectangle.

He can now stack straw on each square, up to a height equal to the length of the side of each square, effectively forming two cubes, and the total volume in cubic feet is a perfect cube.

What was the perimeter of the original rectangle?

[teaser1973]

## Jim Randell 7:50 am

on17 September 2019 Permalink |As presented this puzzle has multiple solutions. However if the perimeter of the original rectangle is required to be a whole number of feet, then only one of the solutions remains.

This Python program finds all the solutions to the puzzle numerically using the [[

`find_zero()`

]] function from theenigma.pylibrary. It runs in 63ms.Run:[ @repl.it ]Solution:The (integer) perimeter of the original rectangle was 48 feet.Allowing non-integer solutions for the perimeter we get four solutions:

Graphically we get a solution when the ellipse

x² + y² – xy = 243intersects the curvex³ + y³ = t³for integer values oft.For positive

xandywe get solutions fort = 16, 17, 18, 19, as shown in the graph below:The original rectangle has sides of length

xandy, so the perimeter of the original rectangle is2(x + y). Onlyt=18gives an integer value for the perimeter (although the values forxandyare not integers). In this case the exact values are:x, y = 12 ± √33.LikeLike

## Jim Randell 8:39 am

on19 September 2019 Permalink |Here is an analytical solution.

Given the equations:

If we consider the product

(x + y)(x² + y² – xy)we have:We also have:

So for a given value of

twe can determine values forx + yandxy, say:These have positive real solutions for

xandywhena² ≥ 4bandb ≥ 0So

tis in the range 16 to 19. And the required perimeter is given byp = 2(x + y) = 2t³ / 243.We can consider the 4 candidate values manually and look for an integer solution, or use a short program:

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