# Compass and straightedge contruction of an equilateral triangle inscribed in a given circle with unknown center

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.

• Try to construct a partial regular hexagon (only some of the vertices are needed). Join the alternate vertices.
– Mick
Mar 17 '15 at 16:45
• What tools are we allowed to use? Can we fold the paper? Can we use a ruler? Mar 17 '15 at 19:24
• @G-man Strict (unmarked) ruler-and-compass construction. From page 21 of the book (which I really must read some time!): "For counting, we consider each use of the ruler to construct a new line as one step, and each use of the compass to construct a new circle as one step. Extending lines previously given or constructed, choosing points at random, and obtaining new points as intersections do not count as separate steps." For some mysterious reason, understood only by moderators, I haven't been allowed to restore the information that this construction is meant to be possible in 7 steps or less. Mar 17 '15 at 19:50
• @CalumGilhooley I don't understand your last statement. Mar 17 '15 at 22:22
• @G-man The OP originally posted an image, directly quoting Hartshorne's statement of the problem. Someone quite correctly and helpfully transcribed this into the highlighted text we see now. In doing so, for some reason they omitted the final part of the statement, which was "(par = 7)". On page 23, Hartshorne explains: "I will sometimes give a par value for a construction, which is the number of steps an experienced constructor would need. By trying harder, you can sometimes succeed with fewer steps." I assumed the omission was unintentional, and tried to correct it by editing the question. Mar 17 '15 at 22:43

Two circles + five lines = 7 steps: 1. 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 $C_0$.

2. 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 $P_3$.

3. Draw a line through points $P_2,P_1$, intersecting $C_0$ at point $P_4$.

4. Draw a line through points $P_1,P_3$, intersecting $C_0$ at point $P_5$.

5. Draw a side $P_5 P_4$ of the inscribed equilateral triangle.

6. Draw a side $P_4 P_0$ of the inscribed equilateral triangle.

7. Draw a side $P_0 P_5$ of the inscribed equilateral triangle.

• Beautiful, just beautiful! Mar 18 '15 at 1:40
• Thank you very much. A really beautiful construction. Mar 18 '15 at 19:25
• @Antonio Lopez Cano: It is indeed. Thank you for bringing the question up, it was a pure joy to solve it. Mar 18 '15 at 22:29
• In Step 1, how do you find a radius $r$ that is sufficiently small without knowing the radius of $C_0$? Jan 8 '18 at 12:50
• @DanielWainfleet: radius of $C_0$ is unknown, but it is assumed that the circle itself can be observed and a reasonably small $r$ can be set. Jan 8 '18 at 13:52

This problem can be found in the online/android/ios puzzle game Euclidea. I came across this thread while researching alternative solutions.

As constructing points doesn't count as a step, the most efficient solution I have found is six steps.

Given circle $c1$ with unknown center:

1. Mark arbitrary points $P1$ and $P2$ on $c1$
2. Draw circle $c2$ with center $P1$ and radius $\overline{P1P2}$
3. Mark point $P3$ at the intersection of $c1$ and $c2$
4. Draw circle $c3$ with center $P2$ and radius $\overline{P2P3}$
5. Mark point $P4$ at the intersection of $c2$ and $c3$
6. Draw circle $c4$ with center $P4$ and radius $\overline{P4P2}$
7. Mark points $P5$ and $P6$ at the intersections of $c3$ and $c4$
8. Draw line through $P5$ and $P2$, intersecting $c1$ at $P7$
9. Draw line through $P6$ and $P2$, intersecting $c1$ at $P8$
10. Draw line segment $\overline{P7P8}$ to complete $\triangle P2P7P8$ If the par score is 7, I'm afraid the best I've managed so far is this bogey 8!

Choose an arbitrary point $A$ on the given circle $\Gamma_0$, and an arbitrary radius, strictly less than the diameter of $\Gamma_0$. All the following circles $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$, $\Gamma_5$ are constructed with this radius.

1. Draw circle $\Gamma_1$, centre $A$, cutting $\Gamma_0$ at points $B$, $B'$.

2. Draw circle $\Gamma_2$, centre $B$, cutting $\Gamma_1$ at point $C$, on same side (of diameter through $A$) as $B'$.

3. Draw circle $\Gamma_3$, centre $B'$, cutting $\Gamma_1$ at point $C'$, on same side (of diameter through $A$) as $B$.

4. Draw circle $\Gamma_4$, centre $C$, cutting $\Gamma_3$ at point $D \ne A$.

5. Draw circle $\Gamma_5$, centre $C'$, cutting $\Gamma_2$ at point $D' \ne A$.

6. Draw $AD$, cutting $\Gamma_0$ at $E$.

7. Draw $AD'$, cutting $\Gamma_0$ at $E'$.

8. Draw $EE'$.

The triangle $AEE'$ is equilateral, and inscribed in $\Gamma_0$. My justification of this construction (in rough, with a blunt pencil, on a very old sheet of graph paper, covered with previous failed attempts) is as follows:

Draw the equilateral triangle $ABC$, and its reflection on the other side of $AB$, whose apex is the other intersection (call it $F$) of $\Gamma_1$ and $\Gamma_2$; and similarly, the equilateral triangle $AB'C'$, and its reflection on the other side of $AB'$, whose apex is the other intersection (call it $F'$) of $\Gamma_1$ and $\Gamma_3$.

With the tangent to $\Gamma_0$ at $A$, the segments $AF$, $AC'$, $AB$, $AB'$, $AC$, $AF'$ make a series of angles: $$\alpha + \left(\frac{\pi}{3} - 2\alpha\right) + 2\alpha + \left(\frac{\pi}{3} - 2\alpha\right) + 2\alpha + \left(\frac{\pi}{3} - 2\alpha\right) + \alpha = \pi.$$ By bisecting the two angles $2\alpha$, we construct two line segments making angles of $\pi/3$ with each other and with the tangent to $\Gamma_0$ at $A$. These suffice to construct the inscribed equilateral triangle.

I hope this sketch of a proof will be enough; it doesn't seem worth labouring, as it didn't make par.

• Alternatively (but still needing 8 steps): ignore the mention of $C'$ in step 3; omit steps 5 and 7. We have now constructed $AE$ in 5 steps. Draw a circle through $E$, centre $A$, cutting $\Gamma_0$ in $E'$; and join $AE'$ and $EE'$. Mar 18 '15 at 1:32

0: Given a circle [c, the black one] (with center NOT given) but radius = r. 1: Start from any point C on c, draw another equal circle (d) cutting c at E and F.

2: Draw 2 other equal circles (e and f, centered at E and F respectively) cutting c at A and B respectively.

3: Join AB, BC, CA.

It takes 3 circles + 3 line = 6 steps.

• How do you construct another circle equal to the given one? (But I like the fact that we have now been offered solutions with 6, 7, 8, and 9 steps.) Mar 18 '15 at 3:53
• @CalumGilhooley The post says “Given a circle, but not given its center”. It has a specific restriction on the center but says nothing about the radius, meaning that the radius should be known (i.e. r). Analyzing from another angle, if a circle with both center and radius unknown, what kind of circle (or question) is that?
– Mick
Mar 18 '15 at 5:38
• That's logical enough, I suppose. On the other hand, how could one effectively know the radius, without having the centre available for setting the compass? The final authority is Euclid, whose Book III, Proposition 1, is: To find the centre of a given circle. This takes him 6 steps, by my estimate; and Hartshorne says (in exercise 2.1) that it can be done in 5; whereas with your assumptions, it could be done in 2. Mar 18 '15 at 11:35
• @CalumGilhooley Note that the assumption depends heavily on "the radius is known".
– Mick
Mar 18 '15 at 18:16

This depends on what "given a circle $C$" means. Suppose you must start with just 2 points $P,Q$ in the plane, neither of them on $C,$ and suppose $C$ is "invisible". And suppose that if any line that is constructible from $P$ and $Q$ intersects $C$ at a point or points, then you can "see" the point(s) of intersection. That is, the point(s) of the intersection are deemed to have been constructed.

Draw lines through $P,$ evenly radially spaced at a radial separation of $\pi2^{-n}$ for $n=1,2,3,...$ until $n$ is large enough that two of these lines intersect $C$ at points $E$ and $F,$ or until one of these lines intersects $C$ at 2 pointes $E$ and $F$. Now that you have 2 points $E,F$ on $C,$ the right bisector of $EF$ will intersect $C$ at points $G,H$. The mid-point of $GH$ is of course the center of $C.$