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?

  • 1
    $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.