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."
Best Answer
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).