10
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Apparently the MSE post is: math.stackexchange.com/a/1044401/43208 $\endgroup$ – Todd Trimble Dec 1 '14 at 4:33
  • 2
    $\begingroup$ Hint: It should, perhaps, be pointed out that this concurrency is just a cleverly disguised version of the (trivial) classic result that the perpendicular bisectors of the sides of a triangle are concurrent at the center of the triangle's circumscribing circle. $\endgroup$ – Robert Bryant Dec 1 '14 at 18:14
  • $\begingroup$ @RobertBryant, thanks for the hint. i see what you are saying. i will post my response as an answer. $\endgroup$ – abel Dec 1 '14 at 18:59
12
$\begingroup$

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.

Given given 5 points $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d},\mathbf{e},$ take the line through the centroid of the triangle determined by some three which is in the direction of the sum of the other two. There are ten ways to do this. CLAIM: all ten meet at a common point.

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.

$\endgroup$
  • $\begingroup$ i like the vector proof. but does this not assume the form for the common point of intersection. even in my earlier proof, i could have argued that by symmetry the form must be as claimed. $\endgroup$ – abel Dec 1 '14 at 19:38
  • $\begingroup$ don't you still need the lengths of $d$ and $e$ to equal so that $d+e$ and $d-e$ are orthogonal. that all five of points are on the circle is still needed. $\endgroup$ – abel Dec 1 '14 at 21:02
  • $\begingroup$ @abel One way to spin it is that the result follows from "in the same direction as the line from the origin to $d+e$" So you can say it just like that and not require equal lengths -OR- do require that and say "in the direction perpendicular to the cord $de$ which, because we required that all lengths are equal, is the same direction as the line from the origin to $d+e$." $\endgroup$ – Aaron Meyerowitz Dec 1 '14 at 21:19
6
$\begingroup$

As this construction was new to me, I wanted to see it. Here are two random examples. The green points are centroids of the triples of points.



Concerning the question, "Can the circle be generalized to any conic?": certainly not straightforwardly.



Update. Here is an illustration of Aaron Meyerowitz's beautiful theorem, for $N=8$, $m=2$, showing coincidence of the $\binom{8}{2}=28$ lines through the $n=6$ hexagon centroids $q$ (green) and (in this case) perpendicular to the line through the two complementary points:



$\endgroup$
  • $\begingroup$ i am afraid it is not the centroid. it is collinear with the center of the circle and the centroid of the five points. it is $(a+b+c+d+e)/3,$ where $a,b,\cdots$ are the five points. for a proof, you can see my answer at MSE. $\endgroup$ – abel Dec 1 '14 at 2:26
  • 1
    $\begingroup$ @abel. Your answer cannot be correct, for two reasons. 1 - if we translate the figure, the intersection point must be translated by the same vector ; this is not satisfied by your point $\frac13(a+b+c+d+e)$. 2 - the construction involves the euclidian structure (perpendicular lines) and the solution cannot be purely affine. $\endgroup$ – Denis Serre Dec 1 '14 at 12:43
  • 2
    $\begingroup$ @abel: Perhaps you shouldn't accept this. I would like to see an answer to your first question, a proof that is not brute-force calculation. $\endgroup$ – Joseph O'Rourke Dec 1 '14 at 13:11
  • 3
    $\begingroup$ If you want to generalize this to more general conic sections, you should adjust what you mean by "orthogonal lines". A conic that is not a parabola will intersect the "line at infinity" in "two points" (possibly a conjugate pair of complex points). There is a unique involution of the line at infinity that has those two points as fixed points. This involution replaces "orthogonality". $\endgroup$ – Jason Starr Dec 1 '14 at 13:54
  • 1
    $\begingroup$ @DenisSerre, my answer is correct. i just need to express in a coordinate free form. let the center of the circle be $O,$ the centroid of the five points $G$ and common points where ten lines intersect be $P.$ then $O,G$ $P$ are collinear and $OG:OP = 3:5$ $\endgroup$ – abel Dec 2 '14 at 3:06
4
$\begingroup$

It is possible to give a syhtnetic proof using the following result:

The perpendiculars dropped from the midpoints of a cyclic quadrilateral to the opposite sides are concurrent. Furthermore, their point of intersection is the symmetric of the center of the circumscribed circle with respect to the centroid of the quadrilateral.

Proof: Let $M,N$ be the midpoints of $AB,CD$ and denote $X,Y$ the projections of $M,N$ on $CD,AB$, respectively. Denote by $Z$ the intersection of $MX,NY$. Denote by $O$ the center of the circle. The perpendicularity condition shows that $OMZN$ is a parallelogram, and thus, $OZ,MN$ have the same midpoint. But the midpoint of $MN$ is the centroid of $ABCD$. It is straightforward to see that the other two perpendiculars also pass through $Z$. $\square$

Returning to the initial configuration of the problem, choose a point $P$ among the $5$ points and consider a homothety $h$ of ratio $3/2$ of center $P$. Denote the rest of the points $A,B,C,D$. The line through the centroid of $PAB$ perpendicular to $CD$ is mapped to the line which passes through the midpoint of $AB$ and is perpendicular on $CD$. Thus the four lines corresponding to triangles which contain $P$ are mapped by $h$ into four concurrent lines like in the result proven above. Because the homothety preserves concurrency, the four lines associated to $P$ are also concurrent. For a point $P$ we denote $P_a,P_b,P_c,P_d$ these concurrent lines.

It is not hard to see that if $P,Q$ are different then $\{P_a,P_b,P_c,P_d\} \cap \{Q_a,Q_b,Q_c,Q_d\}$ has at least two elements, thus the concurrency point is the same for every $4$-family of lines associated to a point on the circle. This shows that the ten lines are concurrent.

Here is a Geogebra configuration.


If you want, you can find the precise location of the intersection point, but for this, the fastest way is using complex numbers, and this has already been done.

$\endgroup$
  • $\begingroup$ is this solution related the Fuss' problem? That is problem number 39 in the 100 Great Problems ... by Dorrie? $\endgroup$ – abel Dec 3 '14 at 0:25
  • $\begingroup$ I do not see the connection. Fuss' problem deals with quadrilaterals which have both inscribed and circumscribed circles. Still, the intersection point in this problem coincides with the intersection point of the two chords determined by the inscribed quadrilateral. Maybe there is a connection... $\endgroup$ – Beni Bogosel Dec 3 '14 at 8:27
3
$\begingroup$

from the hint given by @Robertbryant, i can answer the question without computation. i will fix the coordinate system with the center at the origin with radius one and the point $d = 1$. the other four points have $a, b, c$ and $d.$

let $c$ and $a$ approach $b.$ considering the triangle $abe$ the center is $\frac{2}{3}a + \frac{1}{3}e$ through this point a line orthogonal to $cd = bd$ is drawn. for the triangle $abc$ the the center is $b$ and line orthogonal to $ed$ is drawn. if you look at a triangle similar to $bed$ but the $1/3$ of the size anchored at $b$ and a side parallel to $ed.$ then, the common point of intersection is the orthocenter of this smaller triangle.

$\endgroup$
1
$\begingroup$

This also seems to hold for the orthocenter, nine point circle, and delongchamps point but not the exeter point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.