Berlin Papyrus 6619/Examples/Problem 1
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Problem 1 from Berlin Papyrus $\mathit { 6619 }$
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
- 1972: Richard J. Gillings: Mathematics in the Time of the Pharaohs: $14$: Equations of the First and Second Degree: Equations of the Second Degree
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Squares Without Pythagoras: $11$