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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

$$x_1^2+y_1^2=1$$

$$x_2^2+y_2^2=1$$

$$x_3^2+y_3^2=1$$

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

$$|ab|=\sqrt{2(1-x_1x_2-y_1y_2)}$$

$$|ac|=\sqrt{2(1-x_1x_3-y_1y_3)}$$

$$|bc|=\sqrt{2(1-x_2x_3-y_2y_3)}$$

$$|ad|=\sqrt{1+x_4^2+y_4^2+2(-x_1x_4-y_1y_4)}$$

$$|cd|=\sqrt{1+x_4^2+y_4^2+2(-x_3x_4-y_3y_4)}$$

$$|bd|=\sqrt{1+x_4^2+y_4^2+2(-x_2x_4-y_2y_4)}$$

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.

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    $\begingroup$ 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
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Hints:

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?

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  • $\begingroup$ 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
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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.

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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.

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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 $$(x_4-a)^2+(y_4-b)^2-r^2 =0 \tag 6$$

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  • $\begingroup$ 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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