geometric-construction
Well this may be simple but I am not getting it.
Give a line segment (of length $l$)(and a segment of unit length if you require) how to construct a line of length $l^{1/3}$ with only a straight edge and compass?
I know how to draw a line with length $\sqrt l$ (through similarity process) but am at a loss at the cube root one. Can someone help?
If you have a segment $AB$, place the unit length segment on the line where $AB$ lies, starting with $A$ and in the direction opposite to $B$; let $C$ be the other point of the segment. Now draw a semicircle with diameter $BC$ and the perpendicular to $A$; this line crosses the semicircle in a point $D$. Now $AD$ is the square root of $AB$.
$\triangle BCD$ is a right triangle, like $\triangle ACD$ and $\triangle ABD$; all of these are similar, so you find out that $AC/AD = AD/AB$. But $AC=1$, so $AD = \sqrt{AB}$.
See the drawing below:
Given $A,B$. Find their midpoint $M$:
Find a line $\ell\|AB$ (see below). On $\ell$, mark points $P,Q,R$ with $|PQ|=|QR|=u$. Let $PB$ and $RA$ intersect in $Z$. Then $ZQ$ intersects $AB$ in $M$.
Given a line $\ell_1$, find a distict line parallel to it:
On $\ell_1$, find $C,D$ with $|CD|=u$. Draw a line $\ell_2\ne\ell_1$ through $C$. Find $E\ne C$ on $\ell_2$ with $|DE|=u$. Find $F\ne C$ on $\ell_2$ with $|FE|=u$. Find $G\ne E$ on $\ell_2$ with $|GF|=u$. Then $FG\|\ell_1$.
I got introduced in such stuff by: P.Schreiber, Theorie der geometrischen Konstruktionen, Berlin 1975. I'm sure there's more readable and modern literature available.
One important step is always to make clear which construction steps are available with a given set of instruments. Here we use:
Often one needs to pick random elements (and then show that the finalk result does not depend on the random choices):
Best Answer
In terms of rational root theorem, it's not possible to draw cubed root of 2. However, I have found folks who figured things out using Euclidean geometry.
This one is the simplest: https://demonstrations.wolfram.com/ConstructingTheCubeRootOfTwo/#more
Also, I came up with a hack myself. My approach is clunky and not precise past the thousandths decimal. I made this hack based upon some simple facts that
So if you had an isosceles triangle with the apex 78.09 degrees and the two sides are length 1, then the base should be 1.259.
Step 1: Create an angle that is .46875 degrees Construct a right angle. Trisect the right angle. Take one of the 30 degree angles and bisect it 6 times. You will get an angle that is .46875 degrees.
Step 2: Create an angle that is 5.625 degrees. Construct a right angle. Bisect the right angle 4 times. You will get an angle that is 5.625 degrees
Step 3: Construct an angle that is 72 degrees. Create a Golden Triangle (two sides equal to phi and the base is 1). One of the base angles will be 72 degrees.
Step 4: Join angles from Step 1 - 3 together (Euclid Book 1 proposition 23). You will get an angle that is 78.09 degrees. Make a an isosceles triangle with sides = 1. The base will be very close to the cubed root of 2.