# Constructing an equilateral triangle with a given segment as a side, using a compass whose radius is less than the length of the segment

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 $$a=c$$)?

• 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. – Cameron Buie Feb 6 at 0:35
• 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? – Cameron Buie Feb 6 at 0:39
• Yes that is correct. – Thomas Sawyer Feb 6 at 0:42
• Drat.... In that case, I have no intuition whether this is possible. – Cameron Buie Feb 6 at 0:44

Start by drawing a point $$B'$$ on the $$AB$$ segment, at a distance $$r from $$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$$.
• You can prove that the sum of angles is $180^\circ$ using Euclid's 5th postulate. – Andrei Feb 6 at 1:03
• So the only way to prove that the triangle we constructed is equilateral is to first prove that the sum of angles is $180^\circ$? Or is there another way? – Thomas Sawyer Feb 6 at 1:07
• I'm confused by your question. You know that the interior angles of all triangles is $180$ because that was proven by Euclid very early on. The fact that you are sitting down with limited materials to do constructions doesn't mean that every thing history has proven in the past is magically revoked for no good reason. – fleablood Feb 6 at 1:12
Construct $$60^o$$ angles at the base:
or construct a small triangle $$\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$$: