PROBLEM Here is a surprisingly intriguing challenge posed on Euclidea, a mobile app for Euclidean constructions. I have a geometric construction that achieves the goal but not within the constraints. Could anyone advise on a construction that would fit within these constraints?

Given A point $P$ on a circle $O$ whose center is not given.

Goal Inscribe an equilateral triangle $\Delta{PQR}$ in $\bigcirc{O}$.

Constraints Construct all three sides of the $\Delta{PQR}$ to achieve the goal. Only an unmarked straightedge and a non-rusty collapsible compass (i.e., a compass that, while on paper, can have any radius but which cannot maintain said radius when not on paper) can be used. Use only from the construction steps below a number of steps that together require just five lines in total.

  1. Construct a point: 0 lines (L).
  2. Mark the intersection of two curves with a point: 0L.
  3. Construct a new line (*or line segment or ray): 1L.
  4. Extend a given line segment (*or ray): 1L.
  5. Construct a circle (non-rusty collapsible compass): 1L.
  6. Construct the perpendicular bisector of a line segment: 1L.
  7. Construct a new line perpendicular to an old line: 1L.
  8. Construct an angle bisector: 1L.

Hints from Euclidea A known solution employs five constructions in the order: Circle, circle, perpendicular bisector, perpendicular bisector, and line.

PREVIOUS WORK As mentioned above, I have a construction that matches neither the 5L-constraint nor the hint. The following figure accompanies the six-line construction below, where the given and goal are shown as an inset:

6L construction to this problem

  1. Construct a point $A \ne P$ on circle $\bigcirc{O}$ [0L running total].
  2. Construct $\bigcirc{A(P)}$ centered on point $A$ and with radius $AP$ [1L running total].
    • $\bigcirc{A(P)}$ also intersects $\bigcirc{O}$ at another point $B \ne P$.
  3. Construct $\bigcirc{P(B)}$ [2L running total].
    • $\bigcirc{P(B)}$ also intersects $\bigcirc{A(P)}$ at another point $C \ne B$.
    • Therefore, $\bigcirc{P(C)} = \bigcirc{P(B)}$.
    • Also, $\overline{CP}$ is tangent to $\bigcirc{O}$ at point $P \Longrightarrow m\angle{CPO} = 90^\circ{}$.
  4. Construct $\bigcirc{C(P)}$ [3L running total].
    • $\bigcirc{C(P)} \cap \bigcirc{P(C)} = \lbrace D, E\rbrace$.
    • $\bigcirc{C(P)} \cong \bigcirc{P(C)} \Longrightarrow m\angle{CPD} = m\angle{CPE} = 60^\circ{} \Longrightarrow m\angle{DPE} = 120^\circ{}$.
    • Without actually constructing anything: Let $\overrightarrow{EP} \cap \bigcirc{P(C)} = E' \Longrightarrow$ $m\angle{DPE'} = m\angle{EPE'} - m\angle{DPE} = 180^\circ{} - 120^\circ{} = 60^\circ{}$.
    • Therefore, $\angle{DPE'}$ is the first corner of the desired triangle.
  5. Construct $\overline{DP}$ [4L running total].
    • $\overline{DP} \cap \bigcirc{O} \cap \neg P = Q$.
    • $\overline{PQ}$ is the first side of the desired triangle.
  6. Construct $\overline{DP}$ [5L running total].
    • $\overline{EP} \cap \bigcirc{O} \cap \neg P = R$.
    • $\overline{PR}$ is the second side of the desired triangle.
  7. Construct $\overline{QR}$ [6L running total].
    • $\overline{QR}$ is the last side of the desired triangle. $\blacksquare$
  • $\begingroup$ How do they define "construct an old line"? I am not familiar with the concept. $\endgroup$
    – mathguy
    Aug 26 '16 at 23:49
  • $\begingroup$ I'll clear it in the text: Extending a pre-existing ('old') line in any direction is a 1L move. $\endgroup$
    – PDE
    Aug 26 '16 at 23:50
  • $\begingroup$ Why would one need that? Why can't "we" always construct a line to be exceptionally long the first time? Do they have a maximum length of a segment that can be constructed? (and if so, how is it related to anything else in the problem) $\endgroup$
    – mathguy
    Aug 26 '16 at 23:54
  • $\begingroup$ @mathguy Any lines that the player constructs are infinite. But in some problems (obviously not this one), finite lines are given in the context of a given geometric figure. Feel free to disregard this point. I myself think it does not apply here. In noting that extending old lines costs 1L, I was simply passing along information about how Euclidea counts construction moves. $\endgroup$
    – PDE
    Aug 27 '16 at 0:04
  • 1
    $\begingroup$ Is a collapsible compass one that cannot hold its radius once removed from the surface? Or simply a non-rusty compass? (ETA: Ahh, apparently it's the former.) $\endgroup$
    – Brian Tung
    Aug 27 '16 at 0:21

Using Euclidea's hint, I think this works (barring possible pathological placement of $A$ that I haven't checked) ...

enter image description here

Let the given point be $P$ and the given circle be $p$.

  1. Let $A\neq P$ be a point inside circle $p$. [0 lines]
  2. Construct circle $a$ about $P$ through $A$. [1 line]
    • Let circles $p$ and $a$ meet at points $B_1$ and $B_2$.
  3. Construct circle $b$ about $B_1$ through $P$. [2 lines]
    • Let circles $a$ and $b$ meet at points $C_1$ and $C_2$.
  4. Construct line $c_1$ as the perpendicular bisector of $\overline{C_1B_2}$. [3 lines]
    • Let circle $p$ and line $c_1$ meet at $Q$.
    • Note that $P$ lies on $c_1$.
  5. Construct line $c_2$ as the perpendicular bisector of $\overline{C_2B_2}$. [4 lines]
    • Let circle $p$ and line $c_2$ meet at $R$.
    • Note that $P$ lies on $c_2$.
  6. Construct $\overleftrightarrow{QR}$. [5 lines]
    • Note that $\triangle PQR$ is equilateral, as desired.

For proof, consider this (differently-tinted) sub-figure:

enter image description here

From congruent equilateral triangles $\triangle B_1 P C_1$ and $\triangle B_1 P C_2$, we deduce that $\stackrel{\frown}{B_1 C_1} \cong \stackrel{\frown}{B_1C_2}$, and thus also that $\angle B_1 B_2 C_1 \cong \angle B_1 B_2 C_2$ (as inscribed angles subtending congruent arcs) with common measure $30^\circ$. A trivial angle chase through right triangles $\triangle B_2 M_1 D_1$ and $\triangle B_2 M_2 D_2$ (and $\triangle B_2 M_2 N$, if you like) shows, then, that $\triangle PD_1D_2$ is (equiangular, and therefore) equilateral, so that the bisector of $\angle D_1 P D_2$ is perpendicular to the common chord $\overline{B_1B_2}$ of (in the notation of the construction) circles $p$ and $a$.

We have shown that lines $c_1$ and $c_2$ (1) meet at a $60^\circ$ angle at $P$, and (2) have an angle bisector that passes through the center of circle $p$. Consequently, they are necessarily side-lines of an equilateral triangle inscribed in $p$ with vertex $P$. $\square$

  • 1
    $\begingroup$ I am not convinced. $\widehat{QPR}$ is clearly a $60^\circ$ angle with this construction, but why $PQR$ is equilateral? $\endgroup$ Aug 27 '16 at 1:36
  • 1
    $\begingroup$ @Semiclassical: If $A$ is chosen somewhere on the circle, then it might be the antipode of $P$, which is problematic. $\endgroup$
    – Blue
    Aug 27 '16 at 1:44
  • 1
    $\begingroup$ Playing around with Geogebra bears out that the triangle is indeed equilateral, but like @JackD'Aurizio I can't see why it's so. I also noticed in Geogebra that, regardless of the choice of $A$, there seems to be a common intersection between the circle $b$, the line $c_2$, and the segment $\overline{B_2 C_1}$. $\endgroup$ Aug 27 '16 at 1:51
  • 3
    $\begingroup$ (+1) Oh, ok, it works. The angle between $QP$ and $PB_1$ is the angle formed by two perpendicular bisectors of $B_2 C_1 C_2$, that equals $\widehat{C_2 C_1 B_2}$, that is also the angle between $RP$ and $PB_2$. Sorry for the previous moment of non-belief. But maybe it is better to prove that $PQR$ is actually equilateral in your answer. $\endgroup$ Aug 27 '16 at 1:51
  • 1
    $\begingroup$ @PDE: It is worth noticing that a bit of angle chasing along the previous lines also proves the concurrency spotted by Semiclassical. Anyway, the angle between $QP$ and $PB_1$ is the angle between the perpendicular bisector of $C_1 C_2$ and the perpendicular bisector of $C_1 B_2$, hence it equals $\widehat{C_2 C_1 B_2}=\frac{1}{2}\widehat{C_2 P B_2}=\widehat{RPB_2}$. $\endgroup$ Aug 27 '16 at 2:05

Here is a proof of the construction by @Blue:

enter image description here

  1. $\angle{B_1C_1P} = 60^\circ$ is bisected by $\overline{C_1C_2} \Longrightarrow \angle{C_2C_1P} = 30^\circ$.
  2. Let $x := \angle{B_2C_1P} \Longrightarrow \angle{B_2C_1B_2} = \angle{B_1C_1P} + \angle{B_2C_1P} = 30^\circ + x$.
  3. $\overline{B_2C_1} \perp \overline{PQ}$ and $\mathbf{(2)} \Longrightarrow \angle{PST} = 60^\circ - x \Longrightarrow \angle{SPT} = 30^\circ + x$.
  4. Inscribed Angle Theorem and $\mathbf{(2)} \Longrightarrow \angle{C_2PB_2} = 2\angle{B_2C_1C_2} = 60^\circ + 2x$.
  5. $\overline{PR}$ is the perpendicular bisector of $\overline{B_2C_2}$ and $\mathbf{(4)} \Longrightarrow \angle{B_2PR} = 30^\circ + x$.
  6. $\overline{PR}$ is the perpendicular bisector of $\overline{B_2C_2}$ and $\mathbf{(4)} \Longrightarrow \angle{C_2PR} = 30^\circ + x$.
  7. $\overline{PS} \perp \overline{B_2C_1} \Longrightarrow \angle{C_1SP} = 60^\circ - x$.
  8. From $\mathbf{(3)}$, $\angle{B_1PC_2} = \angle{C_2PS} + \angle{SPT} = \angle{C_2PS} + (30^\circ + x) = 60^\circ$ $\Longrightarrow \angle{C_2PS} = 30^\circ - x$.
  9. By $\mathbf{(6)}$ and $\mathbf{(8)}$, $\angle{QPS} = 60^\circ$.
  10. From $\mathbf{(3)}$ and $\mathbf{(5)}$, $\angle{B_1PO} = \angle{B_2PO} \Longrightarrow \angle{TPS} + \angle{OPQ} = \angle{B_2PR} + \angle{OPR}$ $\Longrightarrow \angle{OPQ} = \angle{OPR}$.
  11. SSA congruence, $\mathbf{(10)}$, $OP = OP$, $OQ = OR \Longrightarrow \Delta{OPQ} \cong \Delta{OPR} \Longrightarrow PQ = PR$.
  12. From $\mathbf{(9)}$ and $\mathbf{(11)}$, $\Delta{PQR}$ is equilateral. QED. $\blacksquare$

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