olymmath

CMO 1988

Chinese Mathematical Olympiad 1988.

Day 1

Problem 1

  • Day 1

Let r1,r2,…,rnr_1,r_2,\dots ,r_n be real numbers. Given nn reals a1,a2,…,ana_1,a_2,\dots ,a_n that are not all equal to 00 , suppose that inequality r1(x1−a1)+r2(x2−a2)+⋯+rn(xn−an)≤x12+x22+⋯+xn2−a12+a22+⋯+an2r_1(x_1-a_1)+ r_2(x_2-a_2)+\dots + r_n(x_n-a_n)\leq\sqrt{x_1^2+ x_2^2+\dots + x_n^2}-\sqrt{a_1^2+a_2^2+\dots +a_n^2} holds for arbitrary reals x1,x2,…,xnx_1,x_2,\dots ,x_n . Find the values of r1,r2,…,rnr_1,r_2,\dots ,r_n .

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Problem 2

  • Day 1

Given two circles C1,C2C_1,C_2 with common center, the radius of C2C_2 is twice the radius of C1C_1 . Quadrilateral A1A2A3A4A_1A_2A_3A_4 is inscribed in C1C_1 . The extension of A4A1A_4A_1 meets C2C_2 at B1B_1 ; the extension of A1A2A_1A_2 meets C2C_2 at B2B_2 ; the extension of A2A3A_2A_3 meets C2C_2 at B3B_3 ; the extension of A3A4A_3A_4 meets C2C_2 at B4B_4 . Prove that P(B1B2B3B4)≥2P(A1A2A3A4)P(B_1B_2B_3B_4)\ge 2P(A_1A_2A_3A_4) , and in what case the equality holds? ( P(X)P(X) denotes the perimeter of quadrilateral XX )

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Problem 3

  • Day 1

Given a finite sequence of real numbers a1,a2,…,ana_1,a_2,\dots ,a_n ( ∗\ast ), we call a segment ak,…,ak+l−1a_k,\dots ,a_{k+l-1} of the sequence ( ∗\ast ) a "long"(Chinese dragon) and aka_k "head" of the "long" if the arithmetic mean of ak,…,ak+l−1a_k,\dots ,a_{k+l-1} is greater than 19881988 . (especially if a single item am>1988a_m>1988 , we still regard ama_m as a "long"). Suppose that there is at least one "long" among the sequence ( ∗\ast ), show that the arithmetic mean of all those items of sequence ( ∗\ast ) that could be "head" of a certain "long" individually is greater than 19881988 .

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Day 2

Problem 4

  • Day 2

(1) Let a,b,ca,b,c be positive real numbers satisfying (a2+b2+c2)2>2(a4+b4+c4)(a^2+b^2+c^2)^2>2(a^4+b^4+c^4) . Prove that a,b,ca,b,c can be the lengths of three sides of a triangle respectively. (2) Let a1,a2,…,ana_1,a_2,\dots ,a_n be nn ( n>3n>3 ) positive real numbers satisfying (a12+a22+⋯+an2)2>(n−1)(a14+a24+⋯+an4)(a_1^2+a_2^2+\dots +a_n^2)^2>(n-1)(a_1^4+ a_2^4+\dots +a_n^4) . Prove that any three of a1,a2,…,ana_1,a_2,\dots ,a_n can be the lengths of three sides of a triangle respectively.

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Problem 5

  • Day 2

Given three tetrahedrons AiBiCiDiA_iB_i C_i D_i ( i=1,2,3i=1,2,3 ), planes αi,βi,γi\alpha _i,\beta _i,\gamma _i ( i=1,2,3i=1,2,3 ) are drawn through Bi,Ci,DiB_i ,C_i ,D_i respectively, and they are perpendicular to edges AiBi,AiCi,AiDiA_i B_i, A_i C_i, A_i D_i ( i=1,2,3i=1,2,3 ) respectively. Suppose that all nine planes αi,βi,γi\alpha _i,\beta _i,\gamma _i ( i=1,2,3i=1,2,3 ) meet at a point EE , and points A1,A2,A3A_1,A_2,A_3 lie on line ll . Determine the intersection (shape and position) of the circumscribed spheres of the three tetrahedrons.

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Problem 6

  • Day 2

Let nn ( n≥3n\ge 3 ) be a natural number. Denote by f(n)f(n) the least natural number by which nn is not divisible (e.g. f(12)=5f(12)=5 ). If f(n)≥3f(n)\ge 3 , we may have f(f(n))f(f(n)) in the same way. Similarly, if f(f(n))≥3f(f(n))\ge 3 , we may have f(f(f(n)))f(f(f(n))) , and so on. If f(f(…f⏟k times(n)… ))=2\underbrace{f(f(\dots f}_{k\text{ times}}(n)\dots ))=2 , we call kk the "length" of nn (also we denote by lnl_n the "length" of nn ). For arbitrary natural number nn ( n≥3n\ge 3 ), find lnl_n with proof.

Answer, solutions and source