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In Euclid's Elements, Book 2, Proposition 14, We are shown how to construct a square from a given rectilinear figure. This allows us to square a rectangle.

Is it possible to do the inverse, creating a rectangle with a given side, with the same area as a given square? enter image description here

I have looked through the rest of Euclid's Elements, and I can't see a proposition doing this. Did I miss one? I also saw references online to the Bolyai-Gerwien Theorem, but I couldn't figure out how to apply it using only a compass and straightedge. I have also tried using the Pythagorean theorem and a 3-4-5 triangle, but it doesn't apply to all squares, so I dropped that.

For a bit of context: I have been plotting different equations on paper, using only a compass and straightedge. With the above mentioned proof, we can plot variations of $\sqrt{x}$, however, we would need to turn a square into a rectangle of width 1 to be able to graph any equations using exponents, like $x^2$.

Note: The construction must be possible with an unmarked straightedge and a compass as the only tools. Preferably something that can be proven to be always true by euclidian methods.

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Let $PC$ be the side length of the given square and $PA$, with $\angle CPA=90^\circ$, the given side length of the desired rectangle. Let $O$ be the intersection of the bisector of $AC$ with $\overleftrightarrow{AP}$. The circle around $O$ through $A$ (and $C$) intersects $AP$ in a second point $B$. Then $PB$ is the other side length of the desired rectangle.

enter image description here

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    $\begingroup$ @Blue Thanks, with the diagram It makes sense. $\endgroup$ – Patrick vD Jan 14 '16 at 21:18
  • $\begingroup$ Yep. Since ACB is a right triangle, ACP and CBP are similar, so AC/CP = CP/PB or CP$^2$ = AC PB. $\endgroup$ – marty cohen Jan 14 '16 at 22:24
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Construct a right triangle whose legs are of lengths $a^2$ (see below) and $b$. $a$ is the side of the square, and $b$ is the given side of the desired rectangle. Lay out the unit length along the $b$-leg starting at the right-angle vertex $O$. From this point $M$ draw a line parallel to the hypotenuse, which intersects the other leg at $N$. The distance $\overline{ON}$ is the length of the other desired side of the rectangle $\frac{a^2}{b}$.

Note: To construct $a^2$ in terms of analytic geometry,
(1) Construct a right triangle with vertices $\{L(-1\mid 0);O(0\mid 0);M(0\mid a)\}$.

(2) Construct the perpendicular to $LM$ at $M$. It will intersect the x-axis at $N(a^2\mid 0)$. $ON$ is of length $a^2$.

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  • $\begingroup$ You'll probably want to explain how to the segment of length $a^2$ in the first place. $\endgroup$ – Blue Jan 14 '16 at 23:08

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