Regular Pentagon Construction by K. Knop

Created with GeoGebra

1. Draw line $OA.\,$ Mark point $R,\,$ the second intersection of the line with the givem circle $O(A).$

2. Draw the circle $R(O)\,$ through $O\,$ centered at $R.\,$ Mark point $N,\,$ one of the intersections of the two circles.

3. Draw the circle $S(O),\,$ through $O\,$ centered at $S.\,$ Mark point $M\,$ - one of the intersections of $S(O)\,$ and $O(A).\,$ Choose the one nearest to $N.$

4. Draw line $MN\,$ and mark point $Q\,$ of its intersection with $OA.$

5. Draw circle $R(Q)\,$ and mark its intersections, say, $B\,$ and $E,\,$ with $O(A).$

6. Draw $BQ\,$ and $EQ\,$ and mark their intersections - $D\,$ and $C\,$ - with $O(A).\,$ $ABDCE\,$ is a regular pentagon.

Assum circle $O(A)\,$ is defined by $x^2+y^2=1.\,$ Then $R(O)\,$ is defined by $(x+1)^2+y^2=1,\,$ and $S(O)\,$ by $(x+2)^2+y^2=4.$

We can find $\displaystyle M=\left(-\frac{1}{4},-\frac{\sqrt{15}}{4}\right)\,$ and $\displaystyle N=\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right).\,$

With these $\displaystyle Q=\left(-\frac{3+\sqrt{5}}{2},0\right)\,$ such that $|QR|=\displaystyle \frac{1+\sqrt{5}}{2}=\varphi,\,$ the Golden ratio.

The rest of the proof is left as an exercise to the reader.

The above problem comes from an uncommon site euclidea, devoted to the problems of Euclidean construction. The site and the problem have been brought to my attention by Konstantin Knop who also generously shared his construction.