# Is this curve the circumference of a circle?

Let $\Gamma$ be a single closed curve with no self-intersections on a plane which satisfies the following condition :

Condition : For any distinct four points $P, Q, R, S$ on $\Gamma$, if the line $PQ$ is orthogonal to the line $RS$, then these four points lie on a circle.

Question : Is $\Gamma$ the circumference of a circle?

Motivation : I like this kind of question such as this question which got me interested in the above question. The answer seems yes, but I'm facing difficulty. Can anyone help?

• It appears that your condition forces $\Gamma$ to be a strictly-convex shape. Further, it appears that any "sharp" vertices or near-vertices are precluded as well... Oct 5 '13 at 15:23
• @abiessu: Yes, I agree with you, but I can't prove this completely. Oct 5 '13 at 16:53
• If you allow (uncountably-many) self-intersections, then you can take $\Gamma$ to be the curve from one endpoint of a line segment to the other, and back again. (In that case, any four distinct points would fail to determine orthogonal lines, so the "on a circle" requirement would be moot.) Shall we assume that the curve has no self-intersections at all?
– Blue
Oct 6 '13 at 0:47
• @Blue: I think we don't need to assume that $\Gamma$ has no self-itersections at all. This is because if there exist some self-intersections, then the situation contradicts the condition as you wrote. This means that the condition itself already rejects the case in which there exist some self-intersections. Oct 6 '13 at 4:13
• @Blue: Thank you for pointing it out. I think you are right. I edited the question. Oct 6 '13 at 7:23

This is a start; the remainder would be explicit formulas in coordinates.

Your curve is compact, there is a pair of points $P,Q \in \Gamma$ whose distance realizes the diameter of the set. Take the perpendicular bisector of the segment $PQ.$ this meets $\Gamma$ in two new points $R,S.$ These lie on a circle, of which $RS$ is a diameter. Now, if you draw a sample segment $PQ,$ and draw a number of circles passing thorough those two points, you will find that the circle of smallest diameter is that for which $PQ$ is already a diameter. It follows that this is the precise circle on which $R,S$ lie. So, without any loss of generality, we may place points of $\Gamma$ on $x,y$-axes at $$P=(0,1), Q = (0,-1), R = (1,0), S = (-1,0).$$ It also follows that no point of $\Gamma$ can lie outside the square $$-1 \leq x \leq 1, \; -1 \leq y \leq 1$$ as it would then be of distance greater than $2$ from one of $P,Q,R,S.$

Finally, given any point $T =(a,b) \in \Gamma,$ say with $a > 0, b > 0$ your four-point rule says that we can find its explicit "reflection" across the $x$-axis by writing down the circle through $T,R,S$ and computing $T_x = (a,b').$ We may then find that reflection across the $y$-axis, call it $T_{xy} = (a', b').$ Or, we could begin by reflecting across the $y$ axis, with $T_y = (a'', b)$ followed by $T_{yx} = (a'', b'').$ My suspicion is that, if $a^2 + b^2 \neq 1,$ the distance between either $TT_{xy}$ or $TT_{yx}$ exceeds $2,$ which is prohibited as that is what we are now calling the distance between $PQ.$ Or, one of $T_x, T_y,T_{xy}, T_{yx}$ lies outside the standard square indicated.

EDIT: with $T =(a,b),$ I get $$T_x = (a, \frac{a^2 -1}{b}),$$ and $$T_y = ( \frac{b^2 -1}{a},b).$$ So, if $a^2 + b^2 = 1,$ both $T_x$ and $T_y$ also lie on the unit circle.

EEEDDDIIIIITTTT: Probably necessary to shrink the square to the overlap of the closed disks around each of $P,Q,R,S$ of radius $2.$ This shape has four corners, one is at $(t,t)$ where $t = \frac{\sqrt 7 - 1}{2} \approx 0.823$

EEEEEEEDDDDDDDDDIIIIIIITTTTTT: At this pont, probably the most efficient thing is to find the point $T$ on $\Gamma$ farthest from the origin; by reflecting or rotating the square, we can require this to be $(x,y)$ with $0 < x,y < 1$ and $x^2 + y^2 > 1;$ this last piece is the assumption. I suspect either $T_{xy}$ or $T_{yx}$ will be farther from the origin than $T,$ contradicting assumption. And that will complete the proof. I worked it out, it does not appear that either $T_{xy}$ or $T_{yx}$ will necessarily be farther from the origin than $T.$ Not a bad idea, though...Note that $P,Q,R,S$ are distance $1$ from the origin; one can play with that.

http://en.wikipedia.org/wiki/The_Rule_of_Four