The perimeter and diagonal length of a rectangle are respectively 100 units and x units. Find, in terms of x, the area of the rectangle.

Let us develop two ideas ‘simultaneously’.

Perimeter and area | Pythagoras |

P = 2 L + 2 B = 100 | L ^{2} + B^{2} = x^{2} |

B = L – 50 | L ^{2} + (L – 50)^{2} = x^{2 } |

A = L × B | L ^{2} + L^{2} – 100L + 2500 = x^{2 } |

A = L ( L – 50) | |

A = L ^{2} – 50L | 2L ^{2}– 100L = x^{2 }– 2500 |

Consequently:

A = ½ (2L

^{2}– 100L)A = ½ ( x

^{2 }– 2500)This solution only works because of a fortuitous coincidence in the coefficients.

In general, we should ‘complete the square.’

2L

^{2}– 100L = x^{2 }– 25002(L

^{2}– 50L) = x^{2 }– 25002(L

^{2}– 50L + 0) = x^{2 }– 25002(L

^{2}– 50L + 625 - 625) = x^{2 }– 25002((L – 25)

^{2}– 625) = x^{2 }– 25002(L – 25)

^{2}– 2 × 625 = x^{2 }– 25002(L – 25)

^{2}– 1250 = x^{2 }– 25002(L – 25)

^{2 }= x^{2 }– 2500 + 12502(L – 25)

^{2 }= x^{2 }– 1250 (L – 25)

^{2 }= ½ x^{2 }– 625L – 25

^{ }= √ (½ x^{2 }– 625)L = 25 + √ (½ x

^{2 }– 625)Having obtained L,

*obviously*we can calculate L^{n}and kL, for various n and k. However simplification would be tedious.In conclusion, competition maths is very different from exam maths. In a competition we are looking for creativity and flexibility. In an exam we want regularity and conformity.