olymmath

APMO 2015

Asian Pacific Mathematics Olympiad 2015.

Paper

Problem 1

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Let ABCA B C be a triangle, and let DD be a point on side BCB C. A line through DD intersects side ABA B at XX and ray ACA C at YY. The circumcircle of triangle BXDB X D intersects the circumcircle ω\omega of triangle ABCA B C again at point Z≠BZ \neq B. The lines ZDZ D and ZYZ Y intersect ω\omega again at VV and WW, respectively. Prove that AB=VWA B=V W.

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Problem 2

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Let S={2,3,4,…}S=\{2,3,4, \ldots\} denote the set of integers that are greater than or equal to 2 . Does there exist a function f:S→Sf: S \rightarrow S such that

f(a)f(b)=f(a2b2) for all a,b∈S with a≠b?f(a) f(b)=f\left(a^{2} b^{2}\right) \text { for all } a, b \in S \text { with } a \neq b ?

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Problem 3

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A sequence of real numbers a0,a1,…a_{0}, a_{1}, \ldots is said to be good if the following three conditions hold. (i) The value of a0a_{0} is a positive integer. (ii) For each non-negative integer ii we have ai+1=2ai+1a_{i+1}=2 a_{i}+1 or ai+1=aiai+2a_{i+1}=\frac{a_{i}}{a_{i}+2}. (iii) There exists a positive integer kk such that ak=2014a_{k}=2014.

Find the smallest positive integer nn such that there exists a good sequence a0,a1,…a_{0}, a_{1}, \ldots of real numbers with the property that an=2014a_{n}=2014.

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Problem 4

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Let nn be a positive integer. Consider 2n2 n distinct lines on the plane, no two of which are parallel. Of the 2n2 n lines, nn are colored blue, the other nn are colored red. Let B\mathcal{B} be the set of all points on the plane that lie on at least one blue line, and R\mathcal{R} the set of all points on the plane that lie on at least one red line. Prove that there exists a circle that intersects B\mathcal{B} in exactly 2n−12 n-1 points, and also intersects R\mathcal{R} in exactly 2n−12 n-1 points.

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Problem 5

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Determine all sequences a0,a1,a2,…a_{0}, a_{1}, a_{2}, \ldots of positive integers with a0≥2015a_{0} \geq 2015 such that for all integers n≥1n \geq 1 : (i) an+2a_{n+2} is divisible by ana_{n}; (ii) ∣sn+1−(n+1)an∣=1\left|s_{n+1}-(n+1) a_{n}\right|=1, where sn+1=an+1−an+an−1−⋯+(−1)n+1a0s_{n+1}=a_{n+1}-a_{n}+a_{n-1}-\cdots+(-1)^{n+1} a_{0}.

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