Construction of Components of Second Bimedial
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Theorem
In the words of Euclid:
- To find medial straight lines commensurable in square only which contain a medial rectangle.
(The Elements: Book $\text{X}$: Proposition $28$)
Proof
Let $\rho$, $\rho \sqrt k$ and $\rho \sqrt \lambda$ be rational straight lines which are commensurable in square only.
Take the mean proportional $\rho \sqrt [4] k$ of $\rho$ and $\rho \sqrt k$, which is medial.
Let $x$ be such that:
- $\rho \sqrt k : \rho \sqrt \lambda = \rho \sqrt [4] k : x$
which gives:
- $x = \dfrac {\rho \sqrt \lambda} {\sqrt [4] k}$
We have that:
- $\rho \sqrt k \frown \!\! - \rho \sqrt \lambda$
where $\frown \!\! -$ denotes commensurability in square only.
Thus:
- $\rho \sqrt [4] k \frown \!\! - x$
From Straight Line Commensurable with Medial Straight Line is Medial it follows that $\rho k^{3/4}$ is also medial.
Then:
\(\ds \rho \sqrt \lambda : x\) | \(=\) | \(\ds \rho \sqrt k : \rho \sqrt [4] k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \rho \sqrt [4] k : \rho\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \cdot \rho \sqrt [4] k\) | \(=\) | \(\ds \rho^2 \sqrt \lambda\) | which is a medial area |
$\blacksquare$
Also see
Historical Note
This proof is Proposition $28$ of Book $\text{X}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text X$. Propositions