12th Chinese 1997 Problem A2 Q 1 is the convex quadrilateral ABCD. P is a point inside ABCD such that the angle PXY is acute for X and Y any two adjacent vertices of ABCD. Derive Q n+1 from Q n by reflecting P in each side of Q n to obtain the vertices of Q n+1 . Which of Q 1 , Q 2 , ... , Q 12 must be similar to Q 1997 ? If Q 1997 is cyclic, which of Q 1 , Q 2 , ... , Q 12 must be cyclic? Solution
: Q 1 , Q 5 , Q 9 must be similar to Q 1997 . The key is to take Q n+1 to have as vertices the feet of the perpendiculars from P to each side of Q n . That makes no essential difference, because it is similar to Q n+1 as defined. So take the feet of the perpendiculars from P to AB, BC, CD, DA to be A', B', C', D' respectively. Then angle PA'A = angle PD'A, so PA'AD' is cyclic, so angle PAB = angle PD'A'. Similarly, if we take A'' to be the foot of the perpendicular from P to A'B' (and similarly for B'' etc), then angle PD'A' = angle PC''D''. Defining A''', A'''' etc similarly, we get angle PAB = angle PA''''B''''. In other words Q is similar to Q 5 , Q 9 , ... , Q 4n+1 , ... . Hence Q 1997 is similar to Q 1 , Q 5 , Q 9 . Note that angle PD'C' = angle PDC, so angle A'D'C' = angle PAB + angle PDC. Clearly, in the general case this will not equal any of the angles DAB, ABC, BCD, CDA. Similarly, angle A''D''C'' = angle PBC + angle PDC, angle A'''D'''C''' = angle PCD + angle PDC. So again, in the general case angles A''D''C'' and A'''D'''C''' will not equal an angle of ABCD. Thus Q 1 , Q 5 and Q 9 are the only quadrilaterals in {Q 1 , Q 2 , Q 3 , ... , Q 12 } which must be similar to Q 1997 . We have angle A''D''C'' = angle PBC + angle PDC and angle C''B''A'' = angle PDA + angle PBA. So angle A''D''C'' + angle C''B''A'' = (angle PBA + angle PBC) + (angle PDA + angle PDC) = angle ABC + angle ADC. Hence Q 1 is cyclic iff Q 3 is cyclic. Hence if Q 1997 is cyclic then all Q odd are cyclic. 12th Chinese 1997 © John Scholes jscholes@kalva.demon.co.uk 22 Dec 2002