Locating three sets of collinear points

Question

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.

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.

Answer 1

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.

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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 $\sigma_C \circ \sigma_B \circ \sigma_A$.

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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 $n$.

Answer 2

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.

(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.

Answer 3

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 $E$.)

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 $C$.

Answer 4

(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.

Answer 5

[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 $P = (\sigma_B \circ \sigma_A)(M) = (\sigma_{A} \circ \sigma_{B_1})(M)$.

Answer 6

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.

(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.

Answer 7

[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 $M=(0,1)$:

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 $T_1 = U(t_1)$.

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.

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.