euclidean-geometry
this relates to a question asked at MSE. i was able to find an answer using complex numbers.
here is the question: there are five points on a circle. take any three points, through the centroid of the three points draw a line orthogonal to the line through the remaining two points. take all ten combinations. it turns out that these ten lines are concurrent.
my questions are: can this fact be proved without any computations? can the circle be generalized to any conic?
This I think is the big clue: " If the line you'll draw and a circle come in contact with each other, then you'll get the coordinates of the point of contact instead of getting 0 as the length of the chord."
If the first line cuts a chord of length $d_1$ then the circle is enclosed in a band centred on the line, ranging from being centred on the line when the circle has diameter $d_1$ to a unit circle offset on either side of the line, and the lines bounding the band are tangents to both these unit circles.
With that in mind, what you should do is draw the third line parallel to the first at a distance of $\frac{d_1}{2}$ from it. If the circle is offset the other side of the line, you then have a second chance, and you draw another parallel line the same distance the other side.
One of these lines will then be either tangent to the circle (if it is centred on the line and has diameter $d_1$), and in this case the coordinates of the point of contact easily allow one to deduce the circle's centre.
Otherwise one of these lines will cut out a chord, and the length of this combined with the original chord length and the distance apart of the parallel lines will allow the radius of the circle to be determined.
But once you know the circle's radius, the chord lengths cut by any two oblique lines allow its centre to be determined.
Very nice problem!
Even more is true for this theorem. Check out this drawing from Arseniy Akopyan wonderful book of Geometry in Figures (Second, extended edition, 2017). On page 65 we find Figure 4.7.29)
In the foreword, Arseniy Akopyan writes
"It is commonly very hard to determine who the author of a certain result is."
He nevertheless provides source for many of the figures in the end of his book. Unfortunately for Figure 4.7.29 he doesn't provide such a reference.
This leads me to the answer: Probably it doesn't have a name (like many "geometry theorems").
Best Answer
Your solution takes two lines, calculates their intersection, and confirms that it is as you claimed. Here is a proof which uses position vectors to show that each line goes through the specified point $$\mathbf{p}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}+\mathbf{e}}{3}.$$ I note at the end that this proof works as well if their are more points (with a slight change) and also in $3$ or more dimensions. The proof certainly works, but I still am not satisfied that it is as clear as it could be.
update I have a more direct proof (at the end) of an even more general result. Unfortunately, as sometimes happens, as the proof improves, the initially surprising result seems less amazing.
The claim is that, given 5 points $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d},\mathbf{e}$ on a circle $\mathbf{x} \cdot \mathbf{x}=r^2$ (i.e. centered at the origin), the line through the point $\mathbf{p}$ and the centroid $$\mathbf{q}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}3$$ of the triangle determined by the first three is perpendicular to the line through the points $\mathbf{d}$ and $\mathbf{e}.$ The first line, $\overline{\mathbf{pq}},$ is in the direction of the vector $\mathbf{d}+\mathbf{e}$ and the second line, $\overline{\mathbf{de}},$ is in the direction of the vector $\mathbf{d}-\mathbf{e}.$ The dot product of these two direction vectors is $\mathbf{d}\cdot \mathbf{d}-\mathbf{e}\cdot \mathbf{e}=r^2-r^2=0.$ Hence they are, indeed, perpendicular.
Generalization:
Given $n+2$ points $\mathbf{a_1},\mathbf{a_2},\cdots,\mathbf{a_{n+2}},$all equally distant from the origin, consider all $\binom{n+2}{2}$ lines determined by choosing $n$ of the points and taking the line through their centroid which intersects and is perpendicular to the line determined by the other $2.$ These lines all pass through the point $$\frac{\mathbf{a_1}+\mathbf{a_2}+\cdots+\mathbf{a_{n+2}}}{n}$$.
update Another proof.
The detail I was missing is that the direction perpendicular to a cord $\overline{\mathbf{de}}$ of a circle is the one from the center of the circle to the centroid $\frac{\mathbf{d}+\mathbf{e}}{2}$ of the cord. Or simply from the center to $\mathbf{d}+\mathbf{e}.$ If we specify the direction in this way we no longer need to mention/require a circle.
Proof: one way is the line through the point $$\mathbf{q}=\frac13(\mathbf{a}+\mathbf{b}+\mathbf{c})$$ in the direction of the vector $\mathbf{d}+\mathbf{e}.$ The general point on this line has the form $$\mathbf{q}+t(\mathbf{d}+\mathbf{e})=\frac13(\mathbf{a}+\mathbf{b}+\mathbf{c})+t(\mathbf{d}+\mathbf{e}).$$ The other $9$ lines are similar with some other partition of the $5$ points. Clearly, the choice $t=\frac13$ shows that the point $$\mathbf{p}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}+\mathbf{e}}{3}$$ is on all $10$ lines.
We could scale up by a factor of $3$ and replace the centroid by the sum of the vertices: The line through the sum of some three, say $\mathbf{a}+\mathbf{b}+\mathbf{c}$, in the direction of the sum of the other two $\mathbf{d}+\mathbf{e}$, passes through the sum of all five.
Theorem: Consider any $N=n+m$ points $\mathbf{a}_1,\mathbf{a}_2,\cdots ,\mathbf{a}_N.$ Consider all $\binom{N}{m}$ lines determined by splitting them into disjoint groups of size $n$ and $m$, taking the centroid $\mathbf{q}$ of the first group and the line through it in the direction of the vector from the origin to the centroid (equivalently, sum) of the second group. Then all these lines are concurrent at the point $$\frac{\mathbf{a_1}+\mathbf{a_2}+\cdots+\mathbf{a_{N}}}{n}.$$
If we scale up by $n$ and replace the centroid by the sum, then the point of concurrence is just $${\mathbf{a_1}+\mathbf{a_2}+\cdots+\mathbf{a_{N}}}.$$
We might need to rule out, or specify what to do, in uninteresting degenerate cases such as when a centroid is at the origin.