olymmath

CMO 1989

Chinese Mathematical Olympiad 1989.

Day 1

Problem 1

  • Day 1

We are given two point sets AA and BB which are both composed of finite disjoint arcs on the unit circle. Moreover, the length of each arc in BB is equal to πm\dfrac{\pi}{m} ( m∈Nm \in \mathbb{N} ). We denote by AjA^j the set obtained by a counterclockwise rotation of AA about the center of the unit circle for jπm\dfrac{j\pi}{m} ( j=1,2,3,…j=1,2,3,\dots ). Show that there exists a natural number kk such that l(Ak∩B)≥12πl(A)l(B)l(A^k\cap B)\ge \dfrac{1}{2\pi}l(A)l(B) .(Here l(X)l(X) denotes the sum of lengths of all disjoint arcs in the point set XX )

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Problem 2

  • Day 1

Let x1,x2,…,xnx_1, x_2, \dots ,x_n ( n≥2n\ge 2 ) be positive real numbers satisfying ∑i=1nxi=1\sum^{n}_{i=1}x_i=1 . Prove that:∑i=1nxi1−xi≥∑i=1nxin−1.\sum^{n}_{i=1}\dfrac{x_i}{\sqrt{1-x_i}}\ge \dfrac{\sum_{i=1}^{n}\sqrt{x_i}}{\sqrt{n-1}}.

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Problem 3

  • Day 1

Let SS be the unit circle in the complex plane (i.e. the set of all complex numbers with their moduli equal to 11 ). We define function f:S→Sf:S\rightarrow S as follow: ∀z∈S\forall z\in S , f(1)(z)=f(z),f(2)(z)=f(f(z)),…,f^{(1)}(z)=f(z), f^{(2)}(z)=f(f(z)), \dots, f(k)(z)=f(f(k−1)(z))(k>1,k∈N),…f^{(k)}(z)=f(f^{(k-1)}(z)) (k>1,k\in \mathbb{N}), \dots We call cc an nn -period-point of ff if cc ( c∈Sc\in S ) and nn ( n∈Nn\in\mathbb{N} ) satisfy: f(1)(c)≠c,f(2)(c)≠c,f(3)(c)≠c,…,f(n−1)(c)≠c,f(n)(c)=cf^{(1)}(c) \not=c, f^{(2)}(c) \not=c, f^{(3)}(c) \not=c, \dots, f^{(n-1)}(c) \not=c, f^{(n)}(c)=c . Suppose that f(z)=zmf(z)=z^m ( z∈S;m>1,m∈Nz\in S; m>1, m\in \mathbb{N} ), find the number of 19891989 -period-point of ff .

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Day 2

Problem 4

  • Day 2

Given a triangle ABCABC , points D,E,FD,E,F lie on sides BC,CA,ABBC,CA,AB respectively. Moreover, the radii of incircles of △AEF,△BFD,△CDE\triangle AEF, \triangle BFD, \triangle CDE are equal to rr . Denote by r0r_0 and RR the radii of incircles of △DEF\triangle DEF and △ABC\triangle ABC respectively. Prove that r+r0=Rr+r_0=R .

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Problem 5

  • Day 2

Given 19891989 points in the space, any three of them are not collinear. We divide these points into 3030 groups such that the numbers of points in these groups are different from each other. Consider those triangles whose vertices are points belong to three different groups among the 3030 . Determine the numbers of points of each group such that the number of such triangles attains a maximum.

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Problem 6

  • Day 2

Find all functions f:(1,+∞)→(1,+∞)f:(1,+\infty) \rightarrow (1,+\infty) that satisfy the following condition: for arbitrary x,y>1x,y>1 and u,v>0u,v>0 , inequality f(xuyv)≤f(x)14uf(y)14vf(x^uy^v)\le f(x)^{\dfrac{1}{4u}}f(y)^{\dfrac{1}{4v}} holds.

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