olymmath

VN-TST 1998

Vietnam Team Selection Test 1998.

Day 1

Problem 1

  • Day 1

Let f(x)f(x) be a real function such that for each positive real cc there exist a polynomial P(x)P(x) (maybe dependent on cc ) such that ∣f(x)−P(x)∣≤c⋅x1998| f(x) - P(x)| \leq c \cdot x^{1998} for all real xx . Prove that ff is a real polynomial.

Answer, solutions and source

Problem 2

  • Day 1

In the plane we are given the circles Γ\Gamma and Δ\Delta tangent to each other and Γ\Gamma contains Δ\Delta . The radius of Γ\Gamma is RR and of Δ\Delta is R2\frac{R}{2} . Prove that for each positive integer n≥3n \geq 3 , the equation: (p(1)−p(n))2=(n−1)2⋅(2⋅(p(1)+p(n))−(n−1)2−8)(p(1) - p(n))^2 = (n-1)^2 \cdot (2 \cdot (p(1) + p(n)) - (n-1)^2 - 8) is the necessary and sufficient condition for nn to exist nn distinct circles Υ1,Υ2,…,Υn\Upsilon_1, \Upsilon_2, \ldots, \Upsilon_n such that all these circles are tangent to Γ\Gamma and Δ\Delta and Υi\Upsilon_i is tangent to Υi+1\Upsilon_{i+1} , and Υ1\Upsilon_1 has radius Rp(1)\frac{R}{p(1)} and Υn\Upsilon_n has radius Rp(n)\frac{R}{p(n)} .

Answer, solutions and source

Problem 3

  • Day 1

Let p(1),p(2),…,p(k)p(1), p(2), \ldots, p(k) be all primes smaller than mm , prove that

∑i=1k1p(i)+1p(i)2>ln(ln(m)).\sum^{k}_{i=1} \frac{1}{p(i)} + \frac{1}{p(i)^2} > ln(ln(m)).

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Day 2

Problem 4

  • Day 2

Find all integer polynomials P(x)P(x) , the highest coefficent is 1 such that: there exist infinitely irrational numbers aa such that p(a)p(a) is a positive integer.

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Problem 5

  • Day 2

Let dd be a positive divisor of 5+199819985 + 1998^{1998} . Prove that d=2⋅x2+2⋅x⋅y+3⋅y2d = 2 \cdot x^2 + 2 \cdot x \cdot y + 3 \cdot y^2 , where x,yx, y are integers if and only if dd is congruent to 3 or 7 (mod20)\pmod{20} .

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Problem 6

  • Day 2

In a conference there are n≥10n \geq 10 people. It is known that:

I. Each person knows at least [n+23]\left[\frac{n+2}{3}\right] other people. II. For each pair of person AA and BB who don't know each other, there exist some people A(1),A(2),…,A(k)A(1), A(2), \ldots, A(k) such that AA knows A(1)A(1) , A(i)A(i) knows A(i+1)A(i+1) and A(k)A(k) knows BB . III. There doesn't exist a Hamilton path.

Prove that: We can divide those people into 2 groups: AA group has a Hamilton cycle, and the other contains of people who don't know each other.

Answer, solutions and source