olymmath

US-TST 2000

USA IMO Team Selection Test 2000.

Other problems

Problem 1

Let a,b,ca, b, c be nonnegative real numbers. Prove that a+b+c3−abc3≤max⁡{(a−b)2,(b−c)2,(c−a)2}.\frac{a+b+c}{3} - \sqrt[3]{abc} \leq \max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}.

Problem 2

Let ABCDABCD be a cyclic quadrilateral and let EE and FF be the feet of perpendiculars from the intersection of diagonals ACAC and BDBD to ABAB and CDCD , respectively. Prove that EFEF is perpendicular to the line through the midpoints of ADAD and BCBC .

Problem 3

Let pp be a prime number. For integers r,sr, s such that rs(r2−s2)rs(r^2 - s^2) is not divisible by pp , let f(r,s)f(r, s) denote the number of integers n∈{1,2,…,p−1}n \in \{1, 2, \ldots, p - 1\} such that {rn/p}\{rn/p\} and {sn/p}\{sn/p\} are either both less than 1/21/2 or both greater than 1/21/2 . Prove that there exists N>0N > 0 such that for p≥Np \geq N and all r,sr, s , ⌈p−13⌉≤f(r,s)≤⌊2(p−1)3⌋.\left\lceil \frac{p-1}{3} \right\rceil \le f(r, s) \le \left\lfloor \frac{2(p-1)}{3} \right\rfloor.

Problem 4

Let nn be a positive integer. Prove that (n0)−1+(n1)−1+⋯+(nn)−1=n+12n+1(21+222+⋯+2n+1n+1).\binom{n}{0}^{-1} + \binom{n}{1}^{-1} + \cdots + \binom{n}{n}^{-1} = \frac{n+1}{2^{n+1}} \left( \frac{2}{1} + \frac{2^2}{2} + \cdots + \frac{2^{n+1}}{n+1} \right).

Problem 5

Let nn be a positive integer. A cornercorner is a finite set SS of ordered nn -tuples of positive integers such that if a1,a2,…,an,b1,b2,…,bna_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n are positive integers with ak≥bka_k \geq b_k for k=1,2,…,nk = 1, 2, \ldots, n and (a1,a2,…,an)∈S(a_1, a_2, \ldots, a_n) \in S , then (b1,b2,…,bn)∈S(b_1, b_2, \ldots, b_n) \in S . Prove that among any infinite collection of corners, there exist two corners, one of which is a subset of the other one.

Problem 6

Let ABCABC be a triangle inscribed in a circle of radius RR , and let PP be a point in the interior of triangle ABCABC . Prove that PABC2+PBCA2+PCAB2≥1R.\frac {PA}{BC^{2}} + \frac {PB}{CA^{2}} + \frac {PC}{AB^{2}}\ge \frac {1}{R}. Alternative formulation: If ABCABC is a triangle with sidelengths BC=aBC=a , CA=bCA=b , AB=cAB=c and circumradius RR , and PP is a point inside the triangle ABCABC , then prove that

PAa2+PBb2+PCc2≥1R\frac {PA}{a^{2}} + \frac {PB}{b^{2}} + \frac {PC}{c^{2}}\ge \frac {1}{R} .