geometric-constructiongeometry
I haven't found a proper solution for this problem, found in Hartshorne's "Geometry: Euclid and Beyond":
(4.3) Given a circle, but not given its center, construct an inscribed equilateral triangle in as few steps as possible.
I managed to construct it in $9$ steps (use of compass or straightedge) but I can't get any lower. Finding circle center takes $5$ of those $9$ uses, and then I need $1$ more to get vertices and $3$ for constructing the triangle.
Square $\square ABCD$, with constructed edge-lines, is inscribed in $\bigcirc O$. (Proof that the quadrilateral is, in fact, a square, is left as an exercise to the reader.)
Edit. Having been asked to elaborate on the square ...
As of Step 2, we know $\triangle P_1 P_2 C$ is equilateral and that $\overline{OA}$ is on the perpendicualr bisector of side $\overline{P_1 P_2}$. Therefore, $\overline{AC}$ is a diameter of $\bigcirc O$, and we have that $\angle ABC$ and $\angle ADC$ (for point $D$ constructed later) are right angles by Thales' Theorem.
As of Step 4, as observed by Jan and Tristan in the comments, $\overline{Q_1 Q_2}$ is a diameter of $\bigcirc{P_1}$, so $\angle Q_1 C Q_2$ is a right angle. Therefore, $\square ABCD$ is at least a rectangle.
Now, define $a := |\overline{OA}|$, so that $|\overline{P_1P_2}| = a\sqrt{3}$ and $|\overline{OQ_1}| = a( 1 + \sqrt{3})$. Since $\angle AOQ_1 = 60^\circ$, if we let $R$ be the foot of the perpendicular from $Q_1$ to $\overleftrightarrow{OA}$, then $|\overline{OR}| = \frac{a}{2}( 1 + \sqrt{3})$ and $$|\overline{Q_1R}| = \frac{a \sqrt{3}}{2}(1+\sqrt{3}) = \frac{a}{2}(3 + \sqrt{3}) = a + |\overline{OR}| = |\overline{CR}|$$ Thus, $\angle Q_1 C R = 45^\circ$ and we may conclude that $\square ABCD$ is a square. $\square$
From @Blue's answer to my Mathematics StackExchange question:
Construct $\bigcirc A$ through $O$.
- Let $P_1$ and $P_2$ be the points where $\bigcirc A$ meets $\bigcirc O$.
Construct $\bigcirc P_1$ through $P_2$.
- Let $C$ be the (other) point where $\bigcirc P_1$ meets $\bigcirc O$.
Construct $\overleftrightarrow{OP_1}$.
- Let $Q_1$ and $Q_2$ be the points where $\overleftrightarrow{OP_1}$ meets $\bigcirc P_1$.
Construct $\overleftrightarrow{CQ_1}$.
- Let $B$ be the point where $\overleftrightarrow{CQ_1}$ meets $\bigcirc O$.
- Note that we have constructed $\overleftrightarrow{BC}$.
Construct $\overleftrightarrow{CQ_2}$.
- Let $D$ be the point where $\overleftrightarrow{CQ_2}$ meets $\bigcirc O$.
- Note that we have constructed $\overleftrightarrow{CD}$.
Construct $\overleftrightarrow{AB}$.
- Construct $\overleftrightarrow{AD}$.
Square $\square ABCD$, with constructed edge-lines, is inscribed in $\bigcirc O$. (Proof that the quadrilateral is, in fact, a square, is left as an exercise to the reader.)
Edit. Having been asked to elaborate on the square ...
As of Step 2, we know $\triangle P_1 P_2 C$ is equilateral and that $\overline{OA}$ is on the perpendicualr bisector of side $\overline{P_1 P_2}$. Therefore, $\overline{AC}$ is a diameter of $\bigcirc O$, and we have that $\angle ABC$ and $\angle ADC$ (for point $D$ constructed later) are right angles by Thales' Theorem.
As of Step 4, as observed by Jan and Tristan in the comments, $\overline{Q_1 Q_2}$ is a diameter of $\bigcirc{P_1}$, so $\angle O_1 C Q_2$ is a right angle. Therefore, $\square ABCD$ is at least a rectangle.
Now, define $a := |\overline{OA}|$, so that $|\overline{P_1P_2}| = a\sqrt{3}$ and $|\overline{OQ_1}| = a( 1 + \sqrt{3})$. Since $\angle AOQ_1 = 60^\circ$, if we let $R$ be the foot of the perpendicular from $Q_1$ to $\overleftrightarrow{OA}$, then $|\overline{OR}| = \frac{a}{2}( 1 + \sqrt{3})$ and $$|\overline{Q_1R}| = \frac{a \sqrt{3}}{2}(1+\sqrt{3}) = \frac{a}{2}(3 + \sqrt{3}) = a + |\overline{OR}| = |\overline{CR}|$$ Thus, $\angle Q_1 C R = 45^\circ$ and we may conclude that $\square ABCD$ is a square. $\square$
Best Answer
Two circles + five lines = 7 steps:
Select an arbitrary point $P_0$ on a given circle $C_0$ and draw a circle $C_1$ with an arbitrary radius $r$, which is small enough to intersect $C_0$.
Draw a circle $C_2$ centered at the intersection point $P_1$ with the same radius $r$ to get the intersection points $P_2$ and $P_3$.
Draw a line through points $P_2,P_1$, intersecting $C_0$ at point $P_4$.
Draw a line through points $P_1,P_3$, intersecting $C_0$ at point $P_5$.
Draw a side $P_5 P_4$ of the inscribed equilateral triangle.
Draw a side $P_4 P_0$ of the inscribed equilateral triangle.
Draw a side $P_0 P_5$ of the inscribed equilateral triangle.