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Problem of the Week
Problem D and Solution
Three Triangles and a Square

Problem

Simon has a rope that is \(200\) cm long. They cut the rope into four pieces so that one piece can be arranged, with its two ends touching, to form a square, and the three remaining pieces can be arranged, with each having its two ends touching, to form three identical equilateral triangles. If all four shapes have integer side lengths, in cm, determine all possibilities for the side lengths of each triangle and the square.

Solution

Let \(x\) represent the side length, in cm, of each equilateral triangle and let \(y\) represent the side length, in cm, of the square.

The perimeter of each figure is the length of the piece of rope used to form it. For each triangle, the length of rope is \(3x\) and for the square the length of rope is \(4y\). The total rope used is \(3(3x)+4y=9x+4y\). But the length of the rope is \(200\) cm. Therefore, \[\begin{aligned} 9x+4y&=200\\ 9x&=200-4y\\ x&=\frac{4(50-y)}{9} \end{aligned}\]

We are given that \(x\) and \(y\) are integers, and the information given in the problem implies \(x\) and \(y\) must be positive. Since both \(x\) and \(y\) are integers, \(4(50-y)\) must be a multiple of \(9\). But \(4\) is not divisible by \(9\), so \(50-y\) must be divisible by \(9\). There are five positive multiples of \(9\) between \(0\) and \(50\), namely \(9\), \(18\), \(27\), \(36\), and \(45\). So \(50-y\) must be equal to \(9\), \(18\), \(27\), \(36\), or \(45\). It follows that \(y\) is equal to \(41\), \(32\), \(23\), \(14\), or \(5\). The corresponding values of \(x\) are computed in the table below.

\(y\) \(4y\) \(200-4y\) \(x=\frac{200-4y}{9}\)
\(41\) \(164\) \(36\) \(4\)
\(32\) \(128\) \(72\) \(8\)
\(23\) \(92\) \(108\) \(12\)
\(14\) \(56\) \(144\) \(16\)
\(5\) \(20\) \(180\) \(20\)

Thus, there are \(5\) possibilities. When the side length of the square is \(41\) cm, the side length of each triangle is \(4\) cm; when the side length of the square is \(32\) cm, the side length of each triangle is \(8\) cm; when the side length of the square is \(23\) cm, the side length of each triangle is \(12\) cm; when the side length of the square is \(14\) cm, the side length of each triangle is \(16\) cm; and when the side length of the square is \(5\) cm, the side length of each triangle is \(20\) cm.