Given 4 coplanar points such that

  • The 4 points do not fall on a circle
  • No 3 of the points fall on a straight line

Is it always true that at least one of the points will be contained in the circle that passes through the remaining 3?

I believe it is probably true, but I have been struggling to find a proof that is not algebraically "messy".

Here's my current thinking: since translations, reflections, rotations, and dilations do not change the essence of the problem, it is okay to make a sequence of these transformations to simplify things a bit. Look for the pair of points that are farthest apart, then do a sequence of transformations to put these 2 points at $(0,0)$ and $(1,0)$. Call these $A$ and $B$.

Now, the remaining distances between any 2 points must be less than 1, which puts a pretty strict limit on where $C$ and $D$ can be located.

I believe that either circle $ABC$ will contain $D$, or circle $ABD$ will contain $C$, or possibly both (after many trials with random points on Geogebra). Since $AB$ is a chord in either circle, and the maximum chord length is $2r$, that means the radius of each circle is at least $\frac{1}{2}$.

I think I have most of the pieces, but I just can't think of how to make the proof "rigorous". Thanks for helping.

Edit: I found a counterexample on Geogebra, where $A=(0,0)$, $B=(1,0)$, $CD < 1$, but neither $C$ nor $D$ is contained in the circle through the remaining 3 points. However, one of $A$ or $B$ was contained in the circle through the other 3.

What I want to be able to do is to take a set of 4 points, and then based on some characteristics of the points (distances, center of mass, etc) be able to say which point(s) will be contained by the circle through the others.

Edit #2: I had previously thought that whichever of the 4 points was closest to the center of mass of the 4 points, i.e. $(\frac{x_1+x_2+x_3+x_4}{4},\frac{y_1+y_2+y_3+y_4}{4})$, would be contained by the circle through the other 3, but I also found a counter example to that on Geogebra.

2 dead ends so far! As Piet Hein says, "Problems worthy of attack prove their worth by hitting back."

  • $\begingroup$ Your idea to put A and B at (0,0) and (0,1) is a good one. Arguments of that type are so common that there is a stock phrase for making one. You say: "Assume without loss of generality that A is at (0,0) and B is at (0,1)." The "without loss of generality" means you are claiming that any other situation can be turned into this one by exploiting some obvious symmetry in the problem. You leave it to the reader to fill in the details. $\endgroup$ – MJD Feb 16 '20 at 15:08

Among all four circles you can make, take the one with the biggest radius.

Say it goes through $ABC$ and $D$ is fourth point and let $A$ and $D$ be on a different sides of line $BC$. Suppose $D$ is outside of the circle and let $AD$ cut the circle at $E$. Draw another circle thorugh $ACD$ with (red) radius $R'$. Then if $CD > CE$ and thus $$ \color{red}{ 2R' ={CD \over \sin \phi}} >{CE \over \sin \phi} =2R$$ a contradiction. So $CE>CD$ and thus $D$ is in the (black circe).

enter image description here

  • $\begingroup$ I wonder how does the proof work in the following case. Assume the angle $ABC$ is obtuse. So is the angle $AEC$. Then the claim $CE<CD$ is generally not valid (btw. I think $CE>CD$ in your answer is a typo). $\endgroup$ – user Jul 1 '20 at 6:41
  • $\begingroup$ Is it better now? @user $\endgroup$ – Aqua Jul 1 '20 at 7:15
  • $\begingroup$ What concerns the typo it is much better. But the proof still does not account for obtuse angle $ABC$, in which case the point $D$ can be outside the black circle but still $CD<CE$. $\endgroup$ – user Jul 1 '20 at 8:38

I have a topological argument. I'm not sure it is what you were looking for, but still I post it. I will not make a rigorous proof to avoid technicities. It enables to understand what situations are possible and what to do to go to one situation from an other.

Let $X$ the space of 4-uplets of points $(A_1, A_2, A_3, A_4)$ pairwise distinct.

When we move continuously the points in the space $X$, two problems might appear :

  • Three of the points become aligned.
  • The four points become cocyclic.

These situations (and only these situations) changes the configuration. In the first case, a circle "is going to the other side of the line". In the second case, some (or several) point is going out of (or in) its corresponding circle.

$A_1, A_2, A_3$ are the points of some triangle $T$. Let us note $L_1, L_2, L_3$ the opposite (infinite) lines of $A_1, A_2, A_3$. There is four possibilities for the point $A_4$, which will determine if $A_1$, $A_2$, $A_3$, $A_4$ are in their corresponding circle (i.e passing by the three other points). I will denote $C_1, .. C_4$ these corresponding circles. When we move continuously $A_4$ :

  1. When $A_4$ is inside the triangle. Then $A_4$ must be in the circle joining $A_1, A_2, A_3$, but neither of $A_1, A_2, A_3$ is inside $C_1, C_2, C_3$ since these circles point towards the opposite direction (for instance it is easily seen that, if $A_4$ is close to $L_1$, $C_1$ will not contain $A_1$, and this is still true for all $A_4$ in the triangle by continuity, since we avoid the "two problems" I mentionned at the beginning).
  2. Else, if $A_4$ is inside the circle, it must be on the other side of some $L_i$. In this case, the circle $C_i$ points towards $A_i$ and is big enough to contain it (the idea is that it would be almost infinite if $A_4$ is close to line $L_i$). The situation for the other circles does not change compared to 1.
  3. Else, if is is outside of the circle, on the same side of $L_i$ and of the two other $L_j$ as 2. Then $C_i$ is too far from $A_i$ to countain it. But now each $C_j$ contains $A_j$. This is easily seen if $A_4$ is far, and it is still true in all the region since in this region we avoid the "two problems".
  4. Else, it is outside of the circle, on the same side of $L_i$ as 2., and on the opposite side of some $L_j$. The situation changes only for the circle $C_j$, which will not contain $A_j$ anymore. Same type of argument.

So there is a limited number of possibilities : there is exactly one or two points which are in the circle formed by the others. In the first case, the point is inside the triangle formed by the others. In the second case, all the points are outside the triangle formed by the others, and the two points contained in their corresponding circles are opposite in the convex quadriedral the four points constitutes.


COMMENT.-The circle through three points being given, if the fourth point is in the blue areas then triangles $P13$, $Q23$ and $R21$ respectively contain points $2,1$ and $3$. It is clear that if the fourth point is sufficiently far away in the white areas the property is verified. Consequently, if there is a counterexample, it must be close enough to the red circle in the white zone. For example for the point $4$ in the attached figure it can be proven with magnitudes in cartesian coordinates that the circle through points $P,1,2$ contains the point $3$

enter image description here


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