Let P be a point inside a continuous closed curve in the plane which does not intersect itself. Show that we can find two points on the curve whose midpoint is P.

**Solution**

*The solution I put up was as follows:*

Take an arbitrary chord AB through P. If P is the midpoint then we are done. So assume it is not. Let A' complete a circuit of the curve starting at A and returning to it. Let the chord through A' and P be A'B'. Let f(A') = A'P/B'P. Then f is a continuous function and f(A_{start}) = 1/f(A_{finish}). So for some point C on the curve f must assume the intermediate value 1, which means that P is the midpoint of this chord.

*However, that is not quite right if the curve is as illustrated above, since defining A' does not unambiguously define B'. Moreover, we cannot simply move A' around the curve - we may get stuck and have to move it back a little way in order to allow B' to complete a circuit. This can probably be rescued, but there is a neater solution due to P Mason (who also pointed out the error above).*

Let C be the curve and C' the curve obtained by rotating C through 180^{o} about P. Let m be a point on C closest to P, and M a point on C furthest from P. Then m must lie inside or on C', and M must lie outside or on C'. Hence C and C' must intersect. Take Q to be any common point. Then the point Q' obtained by rotating Q through 180^{o must also lie on C and C'. Now P is the midpoint of QQ'.
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© John Scholes

jscholes@kalva.demon.co.uk

30 Nov 1999

Last corrected/updated 8 Mar 2003