# Heron of Alexandria/Problems/Rectangles

## Problem

Find two rectangles with integral sides, such that:
the area of the first is $3$ times the area of the second
the perimeter of the second is $3$ times the perimeter of the first.

## Solution

The rectangles are:

$53 \times 54$

and:

$318 \times 3$

## Proof

As can be seen:

 $\ds 53 \times 54$ $=$ $\ds 2862$ $\ds$ $=$ $\ds 3 \times \paren {318 \times 3}$

and:

 $\ds 3 \times \paren {2 \times \paren {53 + 54} }$ $=$ $\ds 642$ $\ds$ $=$ $\ds 2 \times \paren {318 + 3}$

and it is seen that the two rectangles fulfil the given conditions.

Let the given ratio be made general, that is, $n$ rather than $3$.

Let $u, v$ and $x, y$ be the sides of the rectangles.

Then we can write:

 $\ds u + v$ $=$ $\ds n \paren {x + y}$ $\ds x y$ $=$ $\ds n u v$

Thus the general solution is:

 $\ds x$ $=$ $\ds 2 n^3 - 1$ $\ds y$ $=$ $\ds 2 n^3$ $\ds u$ $=$ $\ds n \paren {4 n^3 - 2}$ $\ds v$ $=$ $\ds n$

Setting $n = 3$ gives the solution.

$\blacksquare$