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.