So I have a problem I have been working on. I am trying to show that if I have $4$ points and $3$ of them are on a circle. Then the last point has to be on the circle if we say that the sum of distances from all the points are equal.

What I tried to use coordinate geometry but I do not see a nice way to do it. If we have the 3 points on a circle $a=(x_1,y_1),b=(x_2,y_2),c=(x_2,y_2)$ and the last point $d=(x_4,y_4)$ then we center the circle at $(0,0)$ and for simplicity set $r=1$ so we get




Now we can use this to simplify the distances from each point







Now we also know that $|ab|+|ac|+|ad|=|ab|+|bc|+|bd|=|ac|+|bc|+|bd|=|ad|+|cd|+|bd|$ from the fact that the sum of distances is equal.

Having all this information now I want to conclude that $x_4^2+y_4^2=1$

I know I can eliminate one of the unknown distances from the equalities but that does not seem to help.

Any input, hints would be appreciated.

  • 2
    $\begingroup$ Can you clarify "the sum of the distances from all the points has to be equal"? $\endgroup$ – Akababa Feb 25 '18 at 6:41
  • $\begingroup$ @akababa i wrote it out at the end. If you take a point, sun the distancesfrom the other points, you get S. now you do it for all the other points and you get S as well. $\endgroup$ – Sorfosh Feb 25 '18 at 6:44
  • $\begingroup$ Do you want to write $$|ab|+|ac|+|ad|=|ab|+|bc|+\color{red}{|bd|}=|ac|+|bc|+\color{red}{|cd|}=|ad|+|cd|+|bd| $$? $\endgroup$ – Jaideep Khare Feb 25 '18 at 6:59
  • $\begingroup$ @Sorfosh You should clearify the expression "the sum of the distances from all the points has to equal" exactly where you wrote it, not at the end of a long post. $\endgroup$ – DonAntonio Feb 25 '18 at 9:18


Consider an arbitrary quadrilateral ABCD with sides a, b, c, d and diagonals $d_1$ and $d_2$.

enter image description here

So according to information given $$a+d+d_1=a+b+d_2=b+c+d_1=c+d+d_2$$

So doesn't this appear to be a rectangle on solving these equations?

Also three of the four points are concyclic then would the fourth point of rectangle be on same circle?

  • $\begingroup$ That works but you do not really use the fact that the three points are on a circle. I want to use that fact since I want to generalize it to 4 points on a circle and forcing the 5th to be on the circle. $\endgroup$ – Sorfosh Feb 25 '18 at 19:01
  • $\begingroup$ @Sorfosh I did use the fact. You did not notice I said three of the four points are concyclic that means they lie on same circle $\endgroup$ – Darkrai Feb 26 '18 at 1:37
  • $\begingroup$ @Sorfosh any three points are on a circle $\endgroup$ – Akababa Feb 27 '18 at 3:15

Actually the condition that the sum of lengths is equal implies that the quadrilateral is a rectangle. If your sides are $a,b,c,d$ and diagonals are $p,q$ you have

$$a+b+q=c+d+q=b+c+p=a+d+p$$ $$\implies a+b+c+d+2q=b+c+a+d+2p\implies p=q$$ $$\implies a+b=c+d=b+c=a+d\implies a=c,b=d$$

So the rectangle lies on a circle.


This is false. The condition about equal distances is satisfied if the four points are the vertices of a parallelogram. Choose 3 points on a circle, so that the two chords connecting the middle point to the other two form an acute angle. The point that completes the parallelogram does not lie on the circle.


You can easily answer the question using algebra.

Condider three points; this gives you three rquations $$(x_1-a)^2+(y_1-b)^2=r^2 \tag 1$$ $$(x_2-a)^2+(y_2-b)^2=r^2 \tag 2$$ $$(x_3-a)^2+(y_3-b)^2=r^2 \tag 3$$ Subtract $(1)$ from $(2)$ and $(3)$ to get $$2(x_1-x_2)\,\color{red}{a}+2(y_1-y_2)\,\color{red}{b}=(x_1^2+y_1^2)-(x_2^2+y_2^2)\tag 4$$ $$2(x_1-x_3)\,\color{red}{a}+2(y_1-y_3)\,\color{red}{b}=(x_1^2+y_1^2)-(x_3^2+y_3^2)\tag 5$$ Solve for $(a,b)$ (simple since two linear equations for two unknown variable); when done, use $(1)$ to get $r^2$.

Now use the fourth point and the question becomes : is $$(x_4-a)^2+(y_4-b)^2-r^2 =0 \tag 6$$

  • $\begingroup$ Where are we using the fact that the sum of all distances is equal for all points? $\endgroup$ – Sorfosh Feb 25 '18 at 18:58
  • $\begingroup$ @Sorfosh. Nowhere but everywhere at the same time ! $\endgroup$ – Claude Leibovici Feb 25 '18 at 19:47
  • $\begingroup$ I am a little confused. We can set $a=b=0$, we can force the center to be $(0,0)$ why bother calculating $(a,b)$. Could you add more detail? It seems like you made it as far as i did. I must be misunderstanding. $\endgroup$ – Sorfosh Feb 26 '18 at 17:37

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