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Geometry Difficulty 5.4 AIME, harder Prove it India

Let (a0,a1,a2,...)(a_0, a_1, a_2, ...) and (b0,b1,b2,...)(b_0, b_1, b_2, ...) be two infinite sequences of integers such that
(anan1)(anan2)+(bnbn1)(bnbn2)=0, (a_n - a_{n-1})(a_n - a_{n-2}) + (b_n - b_{n-1})(b_n - b_{n-2}) = 0,
for all integers n2n \ge 2. Prove that there exists a positive integer KK such that
aK+2011=aK+(2011)2011. a_{K+2011} = a_{K+(2011)^{2011}}.

Solution

Consider points Pj=(aj,bj)P_j = (a_j, b_j) in the plane. The slope of the lines PnPn1P_nP_{n-1} and PnPn2P_nP_{n-2} are
r=bnbn1anan1,s=bnbn2anan2, r = \frac{b_n - b_{n-1}}{a_n - a_{n-1}}, \quad s = \frac{b_n - b_{n-2}}{a_n - a_{n-2}},
respectively. The given condition implies that rs=1rs = -1. Hence it follows that the lines PnPn1P_nP_{n-1} and PnPn2P_nP_{n-2} are perpendicular to each other. This implies that PnP_n lies on the circle CnC_n with diameter Pn1Pn2P_{n-1}P_{n-2}. Let f(n)=Pn1Pn22f(n) = |P_{n-1} - P_{n-2}|^2. Then f(n)f(n) is an integer. The observation that PnP_n lies on the circle CnC_n shows that f(n)f(n1)f(n) \le f(n-1). Thus we get a non-increasing sequence of positive integers. This must be constant after certain stage. Thus f(n)f(n) is constant for nNn \ge N, for some positive integer NN. This implies that the diameter of CnC_n is constant for all nNn \ge N. Hence CnC_n are all equal circles for nNn \ge N. This implies that Pn=Pn+2P_n = P_{n+2} for all nNn \ge N. But then an=an+2a_n = a_{n+2} for all nNn \ge N. Hence
aN+2011=aN+(2011)2011. a_{N+2011} = a_{N+(2011)^{2011}}.

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