Let w be a circle and AB a line not intersecting w. Given a point P0 on w, define the sequence P0,P1,… as follows: Pn+1 is the second intersection with w of the line passing through B and the second intersection of the line APn with w. Prove that for a positive integer k, if P0=Pk for some choice of P0, then P0=Pk for any choice of P0. Gheorge Eckstein
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Tags: geometry, circumcircle, incenter, geometry unsolved
13.01.2008 17:41
Lets consider two circles w1 and w2 ortogonal to w with centers in points A and B respectively. I∈w1∩w2. I is exist because line AB doesn't intersect w. Lets consider inversion in point I. We see P′n+1 is getting from P′n by composition of two symmetries: relative to lines w′1 and w′2. w′1 and w′2 are passing through center O′ of w′. ∠P′kO′P′k+1=2∠(w′1,w′2)=α So ∠P′kO′P′k+1 is constant. ∠P′kO′P′0=kα−2nπ=0 So P0=Pk for any sequence P0, P1, ...,Pk which is built in such way.
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13.01.2008 19:15
The solution is very nice, thanks! Could the problem G8 from the IMO 1999 also be solved using inversion: Points A, B, C divide the circumcircle w of the △ABC into three arcs. Let X be a variable point on the arc AB and O1, O2 be the incenters of the △CAX and △CBX. Prove that the circumcircle of the △XO1O2 intersects w in a fixed point.
07.09.2016 14:20
This one(IMO)is too simple just use similarities.
29.01.2024 13:21
We prove, more generally that the statement is true for w any conic and AB a line not intersecting w. Problem: Let w be any conic and AB a line not intersecting w. Given a point P0 on w, define the sequence P0,P1,… as follows: Pn+1 is the second intersection with w of the line passing through B and the second intersection of the line APn with w. Prove that for a positive integer k, if P0=Pk for some choice of P0, then P0=Pk for any choice of P0. Proof: Consider a projective transformation which sends ↔AB to the line at infinity. So, w gets sent to a conic which does not intersect the line at infinity, therefore an ellipse. Now, we can take a affine transformation which makes w a circle. Thus, it suffices to prove it for w a circle, and A and B as points at infinity. Let ℓA,ℓB be two lines which have infinite points as A,B. Now, let Qi be the point on w such that QiPi∥ℓA. Then we have QiPi+1∥ℓB. This implies that ∠(PiQi,Pi+1Qi)=∠(ℓA,ℓB)=α and so Pi+1 is obtained from Pi by a rotation of β=2α at the centre of w. This immediately implies what we desire, since P0=Pk if and only if kβ=2nπ for some integer n. ◻