Say I have a line segment $AB$. I have a compass that can only create a circle with some random radius $r$ that is less than the length of $AB$. (I also have a straight-edge to create lines of arbitrary length.)

In this case, is there a general method for constructing an equilateral triangle with $AB$ as one side, using only Euclid's 5 postulates and common mathematical notions(ex: if $a=b$ and $b=c$ then $a=c$)?

  • $\begingroup$ Oh, dear! I have misremembered what Euclid's fifth postulate actually said! I have deleted my answer for the time being, as I see if it can be salvaged. $\endgroup$ – Cameron Buie Feb 6 '19 at 0:35
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    $\begingroup$ For clarity's sake, let me confirm: your available tools are (1) a straight-edge and (2) a compass of fixed radius $r$ less than the length of $AB$? Put another way, it seems that, while you are allowed an idealized straight edge of arbitrary length, your compass is restricted to radius $r.$ Is that right? $\endgroup$ – Cameron Buie Feb 6 '19 at 0:39
  • $\begingroup$ Yes that is correct. $\endgroup$ – Thomas Sawyer Feb 6 '19 at 0:42
  • $\begingroup$ Drat.... In that case, I have no intuition whether this is possible. $\endgroup$ – Cameron Buie Feb 6 '19 at 0:44

Start by drawing a point $B'$ on the $AB$ segment, at a distance $r<AB$ from $A$, towards $B$. You know how to construct $C'$ say above $AB$, such that $AB'C'$ is an equilateral triangle with side $r$. Note that $\angle B'AC'=\angle BAC'=60^\circ$. Now repeat the same procedure at point $B$: find $A''$ towards $A$ on $AB$, at distance $r$ from $B$, then construct $C''$ so that the triangle $A''BC''$ is equilateral, with $C''$ on the same side of $AB$ as $C'$. Once again, notice that $\angle A''BC''=\angle ABC''=60^\circ$. Now all you need to do is extend $AC'$ and $BC''$ until they meet at a point $C$. Since two of the angles in the $ABC$ triangle are $60^\circ$, you have an equilateral triangle, with side $AB$.

  • $\begingroup$ But how could we prove this without knowing that the sum of the interior angles of a triangle are 180 degrees? Because without that knowledge we could have only an isosceles triangle. $\endgroup$ – Thomas Sawyer Feb 6 '19 at 1:00
  • $\begingroup$ You can prove that the sum of angles is $180^\circ$ using Euclid's 5th postulate. $\endgroup$ – Andrei Feb 6 '19 at 1:03
  • $\begingroup$ So the only way to prove that the triangle we constructed is equilateral is to first prove that the sum of angles is $180^\circ$? Or is there another way? $\endgroup$ – Thomas Sawyer Feb 6 '19 at 1:07
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    $\begingroup$ I'm confused by your question. You know that the interior angles of all triangles is $180$ because that was proven by Euclid very early on. The fact that you are sitting down with limited materials to do constructions doesn't mean that every thing history has proven in the past is magically revoked for no good reason. $\endgroup$ – fleablood Feb 6 '19 at 1:12
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    $\begingroup$ One thing pretty subtle about Euclids elements and geometric constructions is that these magic tools of compass and straightedges aren't actually for making things; they are for verifying it is okay to assume figures of specific dimensions can exist. That is you can find of distance of foo, then you know lines of distance foo will always exist in any place and any orientation. "Contructing" was just Euclids way of saying we know they exist. $\endgroup$ – fleablood Feb 6 '19 at 1:22

Construct $60^o$ angles at the base: enter image description here

or construct a small triangle $\triangle AEG$ with the radius of your compass and then find $C$ as the intersection of the line $AG$ and the line $a$ from the point $B$ parallel to the segment $EG$:

enter image description here


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