29
$\begingroup$

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?

  • 0
    @Silverfish I was toying w a similar problem: given a segment of length k, can we "reconstruct" a unit length? Depends on k. In my case, say k is the golden ratio. Then if I square it and subtract k from its square, I obtain my unit length. But I can't square arbitrary k w/o defining an absolute unit length? Likewise, I can't divide an arbitrary segment of length k by itself without explicitly knowing the unit length? I can bisect 2, trisect 3, but arbitrary k is not possible, such as a segment "defined" as length "pi" relative to a "unit length".2017-03-11

3 Answers 3

53

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:

constructing square root of a line segment

  • 0
    Maybe it should be noted that this is essentially [Euclid's proposition II.14](https://mathcs.clarku.edu/~djoyce/elements/bookII/propII14.html) - not so much [Euclid's proposition III.35](https://mathcs.clarku.edu/~djoyce/elements/bookIII/propIII35.html) - which might be equivalent?2018-08-31
11

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.

  • 1
    @Kaestur: Ahh, yes, that's what I've called the "power of a point" theorem (or, at least, that's the relevant form where the point is inside the circle).2010-07-26
0

take a line AB of 1 unit. draw a line segment BC perpendicular to AB and join CA. take the radius of CA and with centre of compass on A draw an arc cutting the extension of line AB. that point is square root 2.(this is in grade 9 syllabus for us). what MAU answered is also another way.