Junior 2010/2

[tanmoy]

A rectangle and a square have the same area,find,with proof,which one has a greater perimeter

[Labib]

Here's a solution:

Let's assume that the rectangle has a base of length $B$ and a height of length $H$ and the square has sides of length $A$.
Now, $BH = A^2 \Rightarrow B = \frac {A^2}H$
We know, the square of a real number cannot be negative.
So,
$(A-H)^2 = A^2+H^2-2AH \geq 0 $
$\Rightarrow A^2 + H^2 \geq 2AH$
$\Rightarrow \frac {A^2}H + H \geq 2A$
$\Rightarrow B + H \geq 2A$
$\Rightarrow 2(B + H) \geq 4A$

So, the rectangle has a greater perimeter. [Proved]