Show that if there are $4$ points and $3$ of them are on a circle, then the last point has to be on the circle if......

Question

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.

Answer 1

Hints:

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

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?

Answer 2

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.

Answer 3

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.

Answer 4

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