Heron of Alexandria/Problems/Rectangles

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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.

$\blacksquare$


Sources

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