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?

  • 1
    $\begingroup$ 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... $\endgroup$
    – abiessu
    Oct 5 '13 at 15:23
  • $\begingroup$ @abiessu: Yes, I agree with you, but I can't prove this completely. $\endgroup$
    – mathlove
    Oct 5 '13 at 16:53
  • $\begingroup$ 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? $\endgroup$
    – Blue
    Oct 6 '13 at 0:47
  • $\begingroup$ @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. $\endgroup$
    – mathlove
    Oct 6 '13 at 4:13
  • 1
    $\begingroup$ @Blue: Thank you for pointing it out. I think you are right. I edited the question. $\endgroup$
    – mathlove
    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.



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.