By Bin Xiong, Peng Yee Lee

The foreign Mathematical Olympiad (IMO) is a contest for top college scholars. China has taken half within the IMO 21 occasions considering 1985 and has received the pinnacle rating for nations 14 occasions, with a mess of golds for person scholars. The six scholars China has despatched each year have been chosen from 20 to 30 scholars between nearly a hundred thirty scholars who took half within the annual China Mathematical pageant in the course of the iciness months. This quantity of includes a suite of unique issues of recommendations that China used to coach their Olympiad crew within the years from 2009 to 2010. Mathematical Olympiad issues of options for the years 2002–2008 look in an past quantity, Mathematical Olympiad in China.

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T-1 +;-a =fJ a,+ ;-a (n = 1, 2, ... ). Then n+l { a, +;_a } becomes a geometric sequence with the Mathematical Olympiad in China 26 common ratio {3, whose first term is 2 al a2 2 +-a-= a +a +-a-=-~-'-. } is _ pn+l _an+l a, - {3 -a ! ,we have (2) Given p = I, q = a = {3 = _ (n - 1, 2, ... ). ~ . By the expression 6. = p 2 - 4q = 0. } is n + 1 (n = 1, 2, ... ). } is Then 1 2s.. = 2 22 + 2a3 + ... + n+I 2n+l • (4)-@, 1 2 5" 3 n +3 = 2 - = 3 _n+3 2" . zn+l • We finally get S .. China Mathernatir:4l Competition Solution a +{3 = p.

L::B ) . Through point C draw PC II MN , intercepting the circle r at point P. I B is the inner center of DABC. Extend line PI to intercept (1) r at point T. ---... (2) For an arbitrary point Q( #A, T, B) on arc AB (not containing C) , denote the inner centers of D AQ C, DQCB by I1, I 2, respectively. Prove that Q, I1, I 2, T are concyclic. Solution (1) As shown in Fig. 2, join NI, MI. Since PC II MN and P , C, M, N are concyclic, PCMN is an isosceles trapezoid. Therefore, NP = MC, PM = NC. Join AM, CI.

This means that there is a real;. > o such that (A -P)(C -B)= A(C -P)(B -A). c-B· Therefore, B , (A-P) = arg (BC _-A) arg C _ p -- -- -- -- which means that the angle of rotation from PC to PAis equal to that from BC to AB . It follows that P , A , B , C are concyclic. The equality in ® holds only when B , P , D are collinear and Plies on the segment BD. This means that f(P) reaches the minimum when P lies on the circumscribed circle of MBC and P , A , B , C are concyclic. (2) From (1) we know that f(P)mm = BD X CA.