Berlin Papyrus 6619/Examples/Problem 1

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Problem 1 from Berlin Papyrus $\mathit { 6619 }$

If it is said to thee:
the area of a square of $100$ is equal to that of two smaller squares.
The side of one is $\frac 1 2 + \frac 1 4$ the side of the other.
Let me know the sides of the two unknown squares.


Solution

The sides of the two unknown squares are $6$ and $8$.


Modern Proof

Let $x$ be the length of the side of the smaller square of the two.

We have:

\(\ds x^2 + y^2\) \(=\) \(\ds 100\) Pythagoras's Theorem
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac 3 4 y\) Definition of $x$
\(\ds \leadsto \ \ \) \(\ds x^2\) \(=\) \(\ds \dfrac 9 {16} y^2\) squaring both sides
\(\ds \leadsto \ \ \) \(\ds \dfrac 9 {16} y^2 + y^2\) \(=\) \(\ds 100\)
\(\ds \leadsto \ \ \) \(\ds \paren {9 + 16} y^2\) \(=\) \(\ds 1600\) multiplying both sides by $16$
\(\ds \leadsto \ \ \) \(\ds 25 y^2\) \(=\) \(\ds 1600\)
\(\ds \leadsto \ \ \) \(\ds y^2\) \(=\) \(\ds 64\) dividing both sides by $25$
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \pm 8\) taking square root of both sides

As this is the side of an actual square we are investigating here, we can dispose of the negative square root.

Thus we have:

$y = 8$

and so:

$x = \dfrac 3 4 \times 8 = 6$

$\blacksquare$


Historic Proof

The method of solution used is the rule of false position.

Assume that:

\(\ds x\) \(=\) \(\ds \dfrac 1 2 + \dfrac 1 4 = \dfrac 3 4\)
\(\ds y\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds x^2 + y^2\) \(=\) \(\ds 1 + \dfrac 9 {16}\)


But the result should be:

$100 = 64 \times \paren {1 + \dfrac 9 {16} }$

Therefore, our two squares must be $64$ times bigger.

Therefore their sides must be $8$ times bigger.


So the result is:

\(\ds x\) \(=\) \(\ds \dfrac 3 4 \times 8\)
\(\ds \) \(=\) \(\ds 6\)
\(\ds y\) \(=\) \(\ds 1 \times 8\)
\(\ds \) \(=\) \(\ds 8\)
\(\ds \leadsto \ \ \) \(\ds x^2 + y^2\) \(=\) \(\ds 100\)

$\blacksquare$


Sources

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