Euclidean Geometry – Construct 60° Angle Through Point in Four Compass-and-Straightedge Steps

Question

PROBLEM Here is a surprisingly intriguing challenge posed on Euclidea, a mobile app for Euclidean constructions: Construct a 60° angle through both a point $P$ and an external (infinite) line $\mathscr{l}$ using:

1. Only three lines at most, OR
2. Only four elementary Euclidean constructions at most.

HINTS FROM EUCLIDEA The problem is not solved until the line from $P$ to the appropriate point in $\mathscr{l}$ is fully constructed. Also, the variants can be solved by the following ordered sequences of moves:

1. Variant 1: Circle, circle, perpendicular bisector.
2. Variant 2: Circle, circle, circle, line.

DEFINITIONS Only unmarked straightedges and collapsing compasses may be used. The following table shows how moves are scored.

1. Construct a point: 0 lines (L), 0 elementary Euclidean moves (E).
2. Marking the intersection of two curves with a point: 0L, 0E.
3. Construct a new line (segment): 1L, 1E.
4. Construct an old line (segment): 1L, 1E.
5. Construct a circle (non-collapsible compass): 1L, 1E.
6. Construct the perpendicular bisector of a line segment: 1L, 3E.
7. Construct a new line perpendicular to an old line: 1L, 3E.
8. Construct an angle bisector: 1L, 4E.

RESEARCH Constructing a 60° angle involving only $\mathscr{l}$ is well-attested. Unfortunately, there are no questions on Mathematics Stackexchange about the $P$-constrained versions above. A Google search does not reveal anything promising.

BEST ATTEMPT YET The following figure accompanies the construction below, where the given and goal are shown as an inset:

1. Construct $\bigcirc{A}$ centered on a point $A$ (in red) in $\mathscr{l}$ and with radius $AP$ [1L, 1E running total].
2. Let $B$ be the rightmost point where $\bigcirc{A}$ $\cap$ $\mathscr{l}$ [1L, 1E running total].
3. Construct $\bigcirc{B_{big}}$ centered on $B$ and with radius $AB$ [2L, 2E running total].
4. Construct $\bigcirc{B_{small}}$ centered on $B$ and with radius $AP$ [3L, 3E running total].
5. Let $C$ be the uppermost point where $\bigcirc{A}$ $\cap$ $\bigcirc{B_{big}}$ [3L, 3E running total].
6. Construct $\bigcirc{C}$ centered on $C$ and with radius $AC$; $m\angle{ABC} = 60°$ [4L, 4E running total].
7. Let $D$ be the leftmost point where $\bigcirc{C}$ $\cap$ $\bigcirc{B_{small}}$ [4L, 4E running total].
8. Construct $\overline{DP}$ [5L, 5E running total].
9. Let $E$ be the point where $\overline{DP}$ $\cap$ $\mathscr{l}$. $m\angle{PEB} = 60°$, which was what we wanted. [5L, 5E running total].

Answer 1

Variant 1

@Blue has provided a solution, excavated here from the depths of the discussions above:

1. Construct a point $Q \in$ the given external line $\ell$ [0L, 0E].
2. Construct $\bigcirc P$ with radius $PQ$ [1L, 1E].
   • Let point $Q' := \bigcirc P \cap \ell \neq Q$.
3. Construct $\bigcirc Q$ with radius $PQ$ [2L, 2E].
   • Let points $S, S' := \bigcirc P \cap \bigcirc Q$; S is the 'upper' intersection WLOG.
   • $m\angle PQS = 60° \implies m\overset{\mmlToken{mo}{⏜}}{PS\,} = 60°$.
4. Construct $\overline{Q'S'}$ [3L, 3E].
   • Resulting $\angle QQ'S'$ inscribes $\overset{\mmlToken{mo}{⏜}}{PS\,} \implies m\angle QQ'S' = 30°$
5. Construct $\ell'$, the perpendicular bisector of $\overline{Q'S'}$ [4L, 6E].
   • Let $U := \ell' \cap \ell$.
   • Let $V := \ell' \cap \overline{Q'S'}$.
   • By construction, $\angle UVQ'$ is a right angle.
   • Therefore, $m\angle VUQ' = 60°$.
   • Moreover, $\ell'$ perpendicularly bisects a chord $\overline{Q'S'}$ of a circle $\bigcirc P \implies \ell'$ contains a diameter of $\bigcirc P \implies \ell'$ passes through $P$.
   • By construction, $\angle VUQ' = \angle PUQ = 60°$, QEF.

Also, Euclidea gives additional points for constructing all possible versions of each geometric object in question. In this case, two possible angles can be produced. Below for completeness, therefore, is the Variant 1-type construction that produces both angles.

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Variant 2

Variant 2 has been solved by @Blue above. A simple angle-chasing proof of that construction is also above. For what it's worth, though, here is an analytical proof:

1. Set $PQ = radius(\bigcirc P) = radius(\bigcirc Q) := 1$.
   • $m\angle PRS = 60°$
2. Define a right-handed coordinate system.
   • $Q := (0, 0)$.
   • $R := (-1, 0)$.
3. Let $\theta := 2 m\angle PQU$.
   • $P = (cos(2\theta), sin(2\theta))$.
   • $S = (cos(2\theta + 60°), sin(2\theta + 60°))$.
4. Calculate $RS^2 = radius^2(\bigcirc R)$. $$ \begin{align} RS^2 & = (x_S - x_R)^2 + (y_S - y_R)^2 \\ & = (cos(2\theta + 60°) - (-1))^2 + (sin(2\theta + 60°) - (0))^2 \\ & = (cos^2(2\theta + 60°) + 2cos(2\theta + 60°) + 1) + sin^2(2\theta + 60°) \\ & = 2 + 2cos(2\theta + 60°) ) \\ RS^2 & = 4cos^2(\theta + 30°) \\ \end{align} $$
5. Determine the equation of $\bigcirc R$. $$ \begin{align} RS^2 & = (x - x_R)^2 + (y - y_R)^2 \\ 4cos^2(\theta + 30°) & = (x + 1)^2 + y^2 \\ \end{align} $$
6. Determine the equation of $\bigcirc P$. $$ \begin{align} PQ^2 & = (x - x_P)^2 + (y - y_P)^2 \\ 1 & = (x - cos(2\theta))^2 + (y - sin(2\theta))^2 \\ \end{align} $$
7. Find the coordinates of $T$, the other intersection of $\bigcirc P$ and $\bigcirc R$.
   $$ \left\{ \begin{aligned} 4cos^2(\theta + 30°) & = (x_T + 1)^2 + y_T^2 \\ 1 & = (x_T - cos(2\theta))^2 + (y_T - sin(2\theta))^2 \\ \end{aligned} \right. $$ WolframAlpha solves this system as $T := \bigcirc P \cap \bigcirc R = \left(cos(2\theta) - \frac{1}{2}, sin(2\theta) - \frac{\sqrt{3}}{2} \right)$.
8. Find the slope $s$ of $\overline{PT}$. $$ \begin{align} s & = \frac{y_T - y_P}{x_T - x_P} = \frac{\left(sin(2\theta) - \frac{\sqrt{3}}{2}\right) - sin(2\theta)}{\left(cos(2\theta) - \frac{1}{2}\right) - cos(2\theta)} \\ s & = \sqrt{3} \end{align} $$
9. The angle of $\overline{PT}$ with respect to the positive x-axis is $atan(s) = 60°$. QED. $\square$