Given any three distinct points $A,B,C$ and a circle $C(O)$, construct points $D,E,F$ on the circle such that

  1. $A,D,E$ are collinear,
  2. $B,E,F$ are collinear and,
  3. $C,F,D$ are collinear.

One such solution is indicated on the diagram below. I have enough analytic and numerical evidence to indicate that these points exists. In fact, there are two such sets of points, as shown by @coproc below. However I would like a geometric construction.

enter image description here

One idea: My idea was to invert the points $A,B,C$ in the circle to find points $A',B',C'$ respectively. The problems then becomes equivalent to the following problem.

Given three points $A',B',C'$ and a circle $C(O)$. Construct three circles such that

  1. One circle passes through points $A'$ and $O$,
  2. another passes through points $B'$ and $O$,
  3. the third passes through points $C'$ and $O$ and,
  4. the three circles intersect pairwise on $C(O)$.

The required points $D,E,F$ are just the points of intersection of these three circles.

  • $\begingroup$ When E slides around on the circle and D and F are generated by intersecting AE and BE, resp., with the circle, it seems that the angle of line EF is monotonously turning. Hence usually you will get two solutions. $\endgroup$ – coproc Oct 14 '15 at 11:12
  • $\begingroup$ I have a system of equations from which we know the coordinates of $D,E,F$ for given coordinates of $A,B,C$, but the system is a bit messy so I don't know if $D,E,F$ are uniquely determined for any given coordinates of $A,B,C$. What is "enough analytic and numerical evidence" and how did you get it? $\endgroup$ – mathlove Oct 16 '15 at 19:54
  • $\begingroup$ @mathlove, I think that it is pretty clear by the intermediate value theorem that such a points must exist. Consider the two tangent lines to the circle that go through the point $A$. These intersect the circle at two points, $E_1$ and $E_2$. Consider lines $BE_1$ and $BE_2$. These intersect the circle at points $F_1$ and $F_2$. Now consider lines $CF_1$ and $CF_2$. These intersect the circle at points $D_1$ and $D_2$. Points $D_1$ and $F_1$ are on opposite side of tangent line $AE_1$ compared with tangent line $AE_2$. Now consider a general point $X$ and drag it from $E_1$ to $E_2$.... $\endgroup$ – Auslander Oct 17 '15 at 10:05
  • $\begingroup$ @David: I see. Thank you for the explanation. $\endgroup$ – mathlove Oct 19 '15 at 10:22

The operation that takes a point $P$ on the circle and find the other intersection point of the line $(AP)$ with the circle is a real automorphism of the circle.

The composition of $3$ such automorphisms is still a real automorphism. If you identify the circle with $\Bbb P^1(\Bbb R)$, it corresponds to a real homography $t \mapsto \frac {at+b}{ct+d}$.

In theory, given $3$ points on the circle and their images you can determine the parameters $a/d,b/d,c/d,\ldots \in \Bbb P^1(\Bbb R)$ and from then construct the line going through the two fixpoints (the parameters of the line are rational fractions in $a,b,c,d$ so they're a rational fraction in the coordinates of the three image points).

There probably even is a way to get a construction that doesn't depend on which $3$ points you choose on the circle, but at the moment I haven't checked on how to construct that line.

Given a point $A \in \Bbb P^2(\Bbb R)$, call $\sigma_A$ the involution of the circle obtained by taking a point $M$ and intersecting the line $(MA)$ with the circle.

Say an automorphism of the circle is direct if the output points turn in the same direction as the input point and indirect if it's not the case (this given by the sign of the determinant $ad-bc$).

The $\sigma_A$ are indirect involutions so they don't represent all the indirect automorphisms (a space of dimension $3$ while we have only $2$ dimensions by picking $A$). However, any direct automorphism can be given in infinitely many ways as a composition of two $\sigma_{A_i}$.

The fixpoints (possibly over $\Bbb C$) of $\sigma_A \circ \sigma_B$ are clearly the (possibly complex) intersections of $(AB)$ with the circle, so for any $C \in (AB)$ there is a unique point $D$ such that $\sigma_A \circ \sigma_B = \sigma_C \circ \sigma_D$. Constructing $D$ or $C$ from the other one is really easy, for example $C$ is the intersection of $(AB)$ with the line $(\sigma_D(M))( \sigma_A \circ \sigma_B (M))$, for any $M$ on the circle.

Next, we clearly have the simplification $\sigma_A \circ \sigma_A = id$.

So given $4$ points we can simplify $(\sigma_A \circ \sigma_B) \circ (\sigma_C \circ \sigma_D)$ by looking at the intersection of $(AB)$ and $(CD)$ (which is possibly at infinity) and moving both $B$ and $C$ to that point while changing $A$ and $D$ appropriately to $A'$ and $D'$. This gives us $(\sigma_A \circ \sigma_B) \circ (\sigma_C \circ \sigma_D) = \sigma_{A'} \circ \sigma_{D'}$

So now that we know how to represent direct automorphisms of the circle with couples of points and compute compositions, and because there is a direct relation between the fixpoints and the line formed by the points, all we have to do is to compute $\sigma_C \circ \sigma_B \circ \sigma_A \circ \sigma_C \circ \sigma_B \circ \sigma_A$ and simplify it to some $\sigma_X \circ \sigma_Y$. Then $(XY)$ intersects the circle at the two fixpoints $D_1$ and $D_2$ of $\sigma_C \circ \sigma_B \circ \sigma_A$.

enter image description here

Note that this construction should work if you replace the circle with any nondegenerate conic, so ellipses, parabolas, and hyperbolas too. You can also generalize this to an odd number $n$ of points instead of $3$, though the complexity of the construction increases linearly with $n$.

  • $\begingroup$ ingenious. Can you elaborate the argument why for any C on AB there is a D with s_A o s_B = s_C o s_D a bit more? $\endgroup$ – coproc Oct 19 '15 at 20:33
  • $\begingroup$ The space of automorphisms of the circle has dimension $3$, any automorphism of the circle is determined by $3$ points. When looking at the $s_C \circ s_D$ with $C,D \in (AB)$, we restrict to the automorphisms $\sigma$ fixing the intersections of the line with the circle, so we have a space of dimension $2$ (an ordered pair on a line) to parametrize a space of dimension $1$ (the image of any fixed point $M$ on the circle). So we can predict that for any $C$ there is a one-to-one correspondance between $D$ and $\sigma(M)$. Since we need $s_C(s_D(M)) = s(M)$ we must have $s_D(M) = s_C(s(M))$ ... $\endgroup$ – mercio Oct 19 '15 at 20:55
  • $\begingroup$ and $s_C(s(M)) = s_D(M)$, which determines $D$ in terms of $C$ and $s(M)$ and $C$ in terms of $D$ and $s(M)$ by drawing the appropriate lines and intersecting it with $(AB)$. $\endgroup$ – mercio Oct 19 '15 at 20:57
  • $\begingroup$ Very clever. It will probably take me a while to properly work though the solution and the construction. I'll award the bounty this afternoon unless someone finds a more elegant construction. Is it obvious why this only works for an odd number of points? $\endgroup$ – Auslander Oct 20 '15 at 1:34
  • $\begingroup$ Ok, for me one missing link was that the considered circle automorphisms (in homogenous coordinates: [t:h] -> [at+bh:ct+dh]) are determined by 3 point correspondences. Can you help me with one more missing link (at least for me): why are the involutions s_A automorphisms of this kind? Is there a geometric interpretation of the parameter t? $\endgroup$ – coproc Oct 20 '15 at 3:27

This is NOT an answer, but showing why there are two solutions.

Let $D$ be a point on the circle and let $E, F$ be the other intersections of lines $AD, BD$, resp., with the circle. When $D$ is rotated around the circle the points $E, F$ rotate around the circle in opposite direction. (The first image only shows $A, D$ and $E$.)

turning intersections

Hence also the line $EF$ rotates around the circle center in the opposite direction. For any point $C$ outside the circle the line $EF$ will run through $C$ twice during one full rotation: once while $E$ is closer to $C$, once while $F$ is closer to $C$.

rotating EF meets C twice

  • $\begingroup$ Yes, you are right about that. I'll adjust the question to reflect your observation. $\endgroup$ – Auslander Oct 18 '15 at 8:20

[This is just a more step by step recipe based on mercio's solution for reference. It is not meant to take the bounty.]

Basic outline:

  1. construct $B_1$ s.t. $\sigma_B \circ \sigma_A = \sigma_A \circ \sigma_{B_1}$
  2. construct $C_1$ s.t. $\sigma_A \circ \sigma_C = \sigma_{C_1} \circ \sigma_{A}$
  3. intersect line $BC$ with line $B_1C_1$ to get $S$
  4. construct $X$ s.t. $\sigma_{C_1} \circ \sigma_{B_1} = \sigma_{S} \circ \sigma_{X}$
  5. construct $Y$ s.t. $\sigma_{C} \circ \sigma_{B} = \sigma_{Y} \circ \sigma_{S}$
  6. intersect line $XY$ with the circle to get $D_1$ and $D_2$, the two possible solutions for $D$.

After step 2 we have the points $B_1$ and $C_1$ s.t. $$\sigma_A \circ \sigma_C \circ \sigma_B \circ \sigma_A = \sigma_{C_1} \circ \sigma_A \circ \sigma_A \circ \sigma_{B_1} = \sigma_{C_1} \circ \sigma_{B_1} .$$ After step 5 we have the points $X$ and $Y$ s.t. $$ \sigma_{C} \circ \sigma_{B} \circ \sigma_{C_1} \circ \sigma_{B_1} = \sigma_{Y} \circ \sigma_{S} \circ \sigma_{S} \circ \sigma_{X} = \sigma_{Y} \circ \sigma_{X} .$$ Hence $$\sigma_C \circ \sigma_B \circ \sigma_A \circ \sigma_C \circ \sigma_B \circ \sigma_A = \sigma_{C} \circ \sigma_{B} \circ \sigma_{C_1} \circ \sigma_{B_1} = \sigma_{Y} \circ \sigma_{X} .$$ The fixed points of $\sigma_C \circ \sigma_B \circ \sigma_A$ are then also the fixed points of $\sigma_{Y} \circ \sigma_{X}$.

The details for construction steps 1, 2, 4, 5 are already given by mercio. I just give the details for step 1 for completeness:

1.1 choose any point $M$ on the circle and intersect line $AM$ with the circle to get the second intersection point $N = \sigma_A(M)$

1.2 intersect line $BN$ with the circle to get the second intersection point $P = \sigma_B(N) = \sigma_B(\sigma_A(M)) = (\sigma_B \circ \sigma_A)(M)$

1.3 intersect line $AP$ with the circle to get the second intersection point $Q = \sigma_A(P) = \sigma_A^{-1}(P)$

1.4 intersect line $QM$ with line $AB$ to get $B_1$, hence $\sigma_{B_1}(M)=Q=\sigma_A^{-1}(P)$ and $P = (\sigma_B \circ \sigma_A)(M) = (\sigma_{A} \circ \sigma_{B_1})(M)$.

  • $\begingroup$ A helpful overview! $\endgroup$ – Auslander Oct 20 '15 at 5:00
  • $\begingroup$ Even more helpful now! $\endgroup$ – Auslander Oct 20 '15 at 7:51

[this is an addition to mercio's first answer: for my understanding there was still a proof missing, why the involved involutions on the circle like $\sigma_A$ (see mercio's answer or below for the precise definition) can be represented as a function of the form $t \mapsto \frac{at+b}{ct+d}$ in the projective line; so I want to share my results on this; I am not aware of any shorter argument]

A circle can be parametrized using stereographic projection. For our needs here it is easiest to consider the unit circle with center $M=(0,1)$:

enter image description here

The real parameter $t$ parametrizes the circle $x^2 + (y-1)^2 = 1$ by $$t \mapsto U(t) := \left(\frac{2t}{t^2+1},\frac{2}{t^2+1}\right) .$$ For a given point $(x,y)$ on the circle we get back $t$ by $t = \frac{x}{y}$. In this parametrization only the point $P$ is missing. The parametrization can be completed by going to homogenous coordinates $[t:1]$ and assigning $[1:0]$ to $P$.

Now let us consider the special situation when $A$ is on the $y$-axis. The involution $\sigma_A$ maps a point $T_0 = U(t_0)$ on the circle to the other intersection of the line $AT$ with the circle $T_1 = U(t_1)$.

enter image description here

Now how are $t_0$ and $t_1$ related? It turns out that $$t_1 = \frac{y_0-2}{y_0t_0} ,$$ where $y_0$ is the $y$-coordinate of $A$. (It takes some algebra to compute the second intersection of the circle $x^2 + (y-1)^2 = 1$ with the line through $A$ and $T_0$, whose equation is $$ y = \frac{\frac{2}{t_0^2+1}-y_0}{\frac{2t_0}{t_0^2+1}}x+y_0 .)$$ So in the special case of $A$ on the $y$-axis the involution $\sigma_A$ can indeed be represented in the form $\frac{0\cdot t+(y_0-2)}{y_0t+0}=\frac{at+b}{ct+d}$.

For the general case of the position of point $A$ it suffices to check that rotating a point on the circle corresponds to a transformation of parameter $t$ of the same form.

enter image description here

Now here the homogenous coordinates come in handy. From the above figure it can be seen that $$ U([t_1:h]^T) = \left(\begin{array}{cc} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{array}\right) U([t_0:h]^T) = U\left( \left(\begin{array}{cc} \cos\frac{\alpha}{2} & -\sin\frac{\alpha}{2} \\ \sin\frac{\alpha}{2} & \cos\frac{\alpha}{2} \end{array}\right) [t_0:h]^T \right) ,$$ which shows that the transformation of $t_0$ to $t_1$ again is of the expected form. Since also concatentions of functions of the form $t \mapsto \frac{at+b}{ct+d}$ are of this form the - at least for me - missing details seem to be filled.

  • $\begingroup$ Great, that seems to complete the proof at least for those of us for whom such results aren't obvious! $\endgroup$ – Auslander Oct 25 '15 at 23:27
  • $\begingroup$ @David some (minor) details (e.g. why t -> (at+b)/(ct+d) can have only 2 fixed points) are still jumped over, but I also feel that the rough outline is complete now. I enjoyed working on this problem, following mercio's "faint path" and communicating with you. And I think I have learned at lot again. Cheers! $\endgroup$ – coproc Oct 26 '15 at 13:09
  • $\begingroup$ coproc and @mercio, yes, I think the remaining details are all pretty clear. For instance, $f(t)=\frac{at+b}{ct+d}$ can only have $2$ fixed points as $f(t)=t$ is quadratic. Unless I'm mistaken, the proof actually shows that a mapping $f$ has a $2$-cycle, since $f\circ f$ has the $2$ fixed points. Showing that these two points must also be fixed points of $f$ shouldn't be too difficult, and probably also follows from the fact that we're effectively dealing with a quadratic. Might you be able to pass on your emails so that your efforts can be properly acknowledged in case I use these results? $\endgroup$ – Auslander Oct 26 '15 at 23:34
  • $\begingroup$ @David we know that $f=\sigma_C\circ\sigma_B\circ\sigma_A$ has two fixed points (see my first "answer"). These two fixed points are also fixed points of $f\circ f$. Since $f\circ f$ can have at most 2 fixed points, the fixed points of $f$ and $f\circ f$ coincide. $\endgroup$ – coproc Oct 27 '15 at 12:23

(Too long for a comment)

If we can represent the coordinates of $D,E,F$ by given coordinates of $A,B,C$, then we may be able to obtain a geometric construction.

Without loss of generality, we may suppose that the circle is $x^2+y^2=1$ and that $A$ is on the $x$-axis. Let $A(a,0),B(b,c),C(d,e),D(p,q),E(r,s),F(t,u)$ where $p^2+q^2=r^2+s^2=t^2+u^2=1$.

Now we want to represent $p,q,r,s,t,u$ by only $a,b,c,d,e$.

Let us consider in complex plane. So, $A,B,C,D,E,F$ is represented as $a,b+ci,d+ei,p+qi,r+si,t+ui$ respectively.

$$\begin{align}&\text{$A,D,E$ are collinear}\\&\iff \frac{p+qi-a}{r+si-a}=\frac{p-qi-a}{r-si-a}\\&\iff (r+si-a)(p-qi-a)=(p+qi-a)(r-si-a)\\&\iff -rq+sp-sa+aq=-ps+qr-aq+as\\&\iff -rq+sp-sa+aq=0\tag1\end{align}$$

$$\begin{align}&\text{$B,E,F$ are collinear}\\&\iff \frac{r+si-b-ci}{t+ui-b-ci}=\frac{r-si-b+ci}{t-ui-b+ci}\\&\iff (t+ui-b-ci)(r-si-b+ci)=(r+si-b-ci)(t-ui-b+ci)\\&\iff -ts+tc+ur-bu+bs-cr=-ru+rc+st-sb+bu-ct\\&\iff -ts+tc+ur-bu+bs-cr=0\tag 2\end{align}$$

$$\begin{align}&\text{$C,F,D$ are collinear}\\&\iff \frac{p+qi-d-ei}{t+ui-d-ei}=\frac{p-qi-d+ei}{t-ui-d+ei}\\&\iff -tq+te+up-du+dq-ep=0\tag3\end{align}$$

So, all we need is to solve the following system :

$$ \begin{cases} p^2+q^2=1\\ r^2+s^2=1\\ t^2+u^2=1\\ -rq+sp-sa+aq=0\\ -ts+tc+ur-bu+bs-cr=0\\ -tq+te+up-du+dq-ep=0\\ \end{cases} $$

However, it is difficult at least for me to represent $p,q,r,s,t,u$ by $a,b,c,d,e$. (This difficulty may imply the difficulty in obtaining a geometric construction)

Anyway, all I can say is that solving this system tells you at least where $D,E,F$ exist numerically, which may not interest you.

  • $\begingroup$ Yes, a very nice approach! But I anticipated a geometric construction. $\endgroup$ – Auslander Oct 20 '15 at 3:08

This is NOT a solution. However, I would like to share my findings.

(1) Wlog, when comparing the distances of A, B, C from O, we can assume that OA is the shortest. By drawing the circle C(O, OA), we can concentrate on the points A, B’ and C’ instead of the original; where B’ is a point on C(O, OA) and is the “image” of B. The same is true for C’ and C. D, E, F are points to be investigated.

enter image description here

(2) The red line (q) is just any line through B. In Geogebra, if you make q movable (i.e. able to EDIT slide in the direction normal to itself /edit), you will find that A and F are fixed but D and E are moving correspondingly along the circle C(O, OF) and B', C' are moving correspondingly along the circle C(O, OA). Of course, drawing other lines of the same nature through A and C will have the similar effect.

  • $\begingroup$ Thanks for your suggestions. I'm kind of reassured that this is not an easy question given that no answer has been suggested yet! Your approach (i.e. obtaining concentric circles) might simplify the situation somewhat. $\endgroup$ – Auslander Oct 15 '15 at 5:39
  • $\begingroup$ How do you construct B' or C' if D,E,F are not known yet? $\endgroup$ – coproc Oct 18 '15 at 20:18
  • $\begingroup$ How does the rotation of line q around B change the position of D and E? $\endgroup$ – coproc Oct 18 '15 at 20:20
  • $\begingroup$ @coproc For Q1, I just assume that if D, E, F is the solution, then B’ is the intersection of BE and the circle AB’C’. If that is the case, instead of studying the points A, B, C (their distances from O are all different), we can concentrate on A, B’, C’ (their distances from O are all equal) instead. For Q2, you have to draw the diagram according to the rule – “distinguishing a new point and a point of intersection” in Geogebra before you can see how D and E move in connection to the rotation of the red line. In addition, the shape of the green triangle changes accordingly. $\endgroup$ – Mick Oct 19 '15 at 5:02
  • $\begingroup$ My question about the red line q is not about GeoGebra, but how it is related to D and E. If q is any line through B, how would its rotation (around B?) change the position of D? $\endgroup$ – coproc Oct 19 '15 at 6:16

Here is another solution. I wasn't satisfied that I didn't find the characteristics of the automorphism just by using what it does on a few points, so this will remedy this.

We have a real automorphism $f$ of our circle $\mathcal C$ (a map given by polynomial equations with real coefficients). This is the same thing as a real automorphism of $\Bbb P^1(\Bbb R)$ (either by randomly labelling three points $0,1,\infty$ on the circle or by taking any rational parametrization, it really doesn't matter).

Now do an extension of scalars to $\Bbb C$ and tada, you obtained an automorphism $f$ of $\Bbb P^1(\Bbb C)$, so of our good old plane with a point at infinity, such that $f$ preserves angles, orientation, and transforms lines/circles to lines/circles.

We know that our automorphism of $\mathcal C$ reverses its orientation so $f$ is a conjugate in $Aut(\Bbb P^1(\Bbb C))$ of multiplication by a real negative number (just conjugate with an automorphism sending the circle to the real line and who sends the two unknown fixpoints to $0$ and $\infty$)
In particular, if $M$ is any point in $\Bbb P^1(\Bbb C)$, if we could compute $f(M)$ and $f(f(M))$ then the circle going through $M$, $f(M)$, $f(f(M))$ will also go through the two fixpoints. Because this is the case when $f$ is multiplication by a negative real number.

So we'll do just that with $M = \infty$, now our goal is simply to construct $f(\infty)$ and $f(f(\infty))$, draw the line between them (a line is a circle going through $\infty$), and intersect it with our circle.

To find $f(\infty)$ we have to find the image of two straight lines and intersect them. Pick a point $X \in \mathcal C$ and let $Y \in \mathcal C$ be its diameter $(XY)$ . We compute $f(X)$ and $f(Y)$. Since $(XY)$ intersects $\mathcal C$ at right angles, so does $f((XY))$. The center of that circle then must be at the intersection of the tangents to $\mathcal C$ at $f(X)$ and $f(Y)$, so we can construct the circle $f((XY))$.
Now do the same from another point $X' \in \mathcal C$ and draw the circle $f((X'Y'))$.

If $O$ is the center of $\mathcal C$, those two circles have to intersect at $f(O)$ and $f(\infty)$ (because both $O$ and $\infty$ are on $(XY)$ and $(X'Y')$). Additionally $f(O)$ must be outside $\mathcal C$ and $f(\infty)$ has to be inside $\mathcal C$ (multiplication by a negative real number switches the bottom and top half-planes)

Next we want to compute $f(f(\infty))$ to do that we compute $f(f(X))$ and $f(f(Y))$ and construct the image of the circle $(X,Y,f(\infty),f(O))$ (remember that they intersect $\mathcal C$ at right angles). The line $(Of(\infty))$ intersects $\mathcal C$ at some point $Z$, compute $f(Z)$ and draw the circle $(f(Z)f(O)f(\infty))$. $f(f(\infty))$ has to be at the intersection of those two circles (and outside $\mathcal C$)

Now draw the line $f(\infty)f(f(\infty))$ and you're done.

enter image description here

(all the lines and points used to construct the images by $f$ are not shown)

And I'm glad that I didn't have to do even more witchcraft to have two circles that look like they don't intersect have them intersect over $\Bbb C$ anyway and have them show up in our complex plane.

  • $\begingroup$ unfortunately this doesn't work with ellipses/parabolas/etc $\endgroup$ – mercio Oct 21 '15 at 18:27
  • $\begingroup$ Nice! I still enjoyed the algebraic/geometric trickery of your first solution, and its generality. $\endgroup$ – Auslander Oct 21 '15 at 22:32

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