Maclaurin 2020 Q3

The three altitudes of a right-angled triangle are the two sides around the right angle and the perpendicular from the right angle to the hypotenuse, which is clearly the shortest.

Let the lengths of the three sides in cm be \(a\), \(b\), \(c\), where \(c\) is the hypotenuse.

From (a) we have \(a + b = 5 - c\), which, when squared, gives \(a^2 + 2ab + b^2 = 25 - 10c + c^2\).

Substituting from (b) and (c), this becomes \(c^2 + 2c = 25 - 10c + c^2\), so \(12c = 25\) and \(c = \tfrac{25}{12}\).

Now we know that \(a + b = \dfrac{35}{12}\) and \(ab = \dfrac{25}{12}\).

Hence \(a + \dfrac{25}{12a} = \dfrac{35}{12}\) and so \(12a^2 - 35a + 25 = 0\), which has solutions \(a = \tfrac{5}{3}\) or \(\tfrac{5}{4}\).

Finally the corresponding values of \(b\) are \(\tfrac{5}{4}\) and \(\tfrac{5}{3}\). Hence there is a unique triangle with sides \(\tfrac{5}{3}\) cm, \(\tfrac{5}{4}\) cm and \(\tfrac{25}{12}\) cm.

An alternative approach uses \(a + b + ab = 5\) to obtain \(a = \dfrac{5 - b}{1 + b}\).

Hence \((\tfrac{5-b}{1+b})^2(b^2 - 1) = b^2\), so we have \((5 - b)^2(b - 1) = b^2(1 + b)\) and this, when expanded, gives \(12b^2 - 35b + 25 = 0\) and so \((3b - 5)(4b - 5) = 0\).

Hence \(b = \tfrac{5}{3}\) or \(\tfrac{5}{4}\) and we find \(a\) and \(c\) as before.

Since this argument is not obviously reversible, it is necessary to check that these lengths satisfy the requirements of the problem.