## Wednesday, June 29, 2011

### Math & Logic Challenge

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 L2 + B2 = x2 B = L – 50 L2 + (L – 50)2 = x2 A = L × B L2 + L2 – 100L + 2500 = x2 A = L ( L – 50) A = L2 – 50L 2L2– 100L = x2 – 2500

Consequently:
A = ½ (2L2– 100L)
A = ½ ( x2 – 2500)

This solution only works because of a fortuitous coincidence in the coefficients.
In general, we should ‘complete the square.’

2L2– 100L = x2 – 2500
2(L2– 50L) = x2 – 2500
2(L2– 50L + 0) = x2 – 2500
2(L2– 50L + 625 - 625) = x2 – 2500
2((L – 25)2– 625) = x2 – 2500
2(L – 25)2– 2 × 625 = x2 – 2500
2(L – 25)2– 1250 = x2 – 2500
2(L – 25)2 = x2 – 2500 + 1250
2(L – 25)2 = x2 – 1250
(L – 25)2 = ½ x2 – 625
L – 25 = (½ x2 – 625)
L = 25 + (½ x2 – 625)

Having obtained L, obviously we can calculate Ln 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.