Compass-and-straightedge construction of the square root of a given line?

Question

Given

1. A straight line of arbitrary length
2. The ability to construct a straight line in any direction from any starting point with the "unit length", or the length whose square root of its magnitude yields its own magnitude.

Is there a way to geometrically construct (using only a compass and straightedge) the a line with the length of the square root of the arbitrary-lengthed line? What is the mathematical basis?

Also, why can't this be done without the unit line length?

Answer 1

Without the unit-length segment--that is, without something to compare the first segment to--its length is entirely arbitrary, so can't be valued, so there's no value of which to take the square root.

Let the given segment (with length x) be AB and let point C be on ray AB such that BC = 1. Construct the midpoint M of segment AC, construct the circle with center M passing through A, construct the line perpendicular to AB through B, and let D be one of the intersections of that line with the circle centered at M (call the other intersection E). BD = sqrt(x).

AC and DE are chords of the circle intersecting at B, so by the power of a point theorem, AB * BC = DB * BE, so x * 1 = x = DB * BE. Since DE is perpendicular to AC and AC is a diameter of the circle, AC bisects DE and DB = BE, so x = DB^2 or DB = sqrt(x).

edit: this is a special case of the more general geometric-mean construction. Given two lengths AB and BC (arranged as above), the above construction produces the length BD = sqrt(AB * BC), which is the geometric mean of AB and BC.

Answer 2

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:

Answer 3

This answer focuses on OP's last question about the unit-length line.

I would answer the rest of the question exactly as mau. It can be seen that in mau's drawing, the unit length segment is a necessary element, and without that segment this construction is not possible. Now I add a small point here:

If we don't have the unit length segment, we may still be able to complete the construction if we know the length whose square root we are seeking. The rationale is that the value of the length contains information. It implies the relative size of a segment with respect to the unit length. So it is useful!

In particular, in the case where the said length is an integer, we are able to use that information to construct a unit length segment (how?). Then we can apply mau's method.