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I have a construction from the game Euclidea, puzzle 4.2:

enter image description here

The puzzle is given point $A$ and line $\overleftrightarrow{BC}$ (just the line -- neither point is given), construct a 60 degree angle with the line through the point (shown in orange). I came up with this construction to satisfy a goal in the game for performing the construction with a minimum number of elementary steps.

The construction is:

  1. Draw $\odot A$, of arbitrary radius large enough to intersect $\overleftrightarrow{BC}$.
  2. Draw $\odot C$.
  3. Draw $\odot D$.
  4. Adding $\overleftrightarrow{EA}$ forms a 60 degree $\angle ABC$, as if by magic.

What's the proof?

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    $\begingroup$ How is circle D constructed? $\endgroup$ – fleablood Jan 31 at 4:32
  • $\begingroup$ @fleablood Circle D is centered on the intersection of circles A and C, through the intersection of circle A with line BC. I've updated the image to label that later intersection F. $\endgroup$ – Phil Frost Jan 31 at 15:18
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As a convention, let's label the circles with two points - in order, the center and one point on the circle. The steps of the construction, with this in mind:

1) Arbitrary large enough circle centered at $A$. Let $C$ be the intersection of this circle with the given line $\ell$.

2) Circle $CA$. Let $D$ be the intersection of circles $AC$ and $CA$, on the opposite side of $\ell$ from $A$. Let $F$ be the intersection of circle $CA$ and $\ell$.

3) Circle $DF$. Let $E$ be the intersection of circle $DF$ and circle $AC$, outside circle $CA$.

4) Line $EA$ forms the desired $60^\circ$ angle with $\ell$.

Yes, I named a point that you didn't bother naming. We used it, so we should name it.

So, why does this work? Note that $\triangle ACD$ is equilateral, so $\angle CDA=60^\circ$. As such, a $60^\circ$ rotation $\rho$ around $D$ takes $C$ to $A$. Because of that, it also takes circle $CD$ to circle $AD$. Taking the intersections of these circles with circle $DF$ (fixed under the rotation, because its center is fixed), $\rho(F)=E$. Then $\rho(\ell)=\rho(\overleftrightarrow{CF})=\overleftrightarrow{\rho(C)\rho(F)}=\overleftrightarrow{AE}$. The angle between a line and its image under the $60^\circ$ rotation $\rho$ must be $60^\circ$. Done.

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  • $\begingroup$ I updated the image to label F for clarity $\endgroup$ – Phil Frost Jan 31 at 15:14

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