olymmath

BG-MO 1989

Bulgarian Mathematical Olympiad 1989.

Other problems

Problem 1

In triangle ABCABC , point OO is the center of the excircle touching the side BCBC , while the other two excircles touch the sides ABAB and ACAC at points MM and NN respectively. A line through OO perpendicular to MNMN intersects the line BCBC at PP . Determine the ratio AB/ACAB/AC , given that the ratio of the area of △ABC\triangle ABC to the area of △MNP\triangle MNP is 2R/r2R/r , where RR is the circumradius and rr the inradius of △ABC\triangle ABC .

Problem 2

Prove that the sequence (an)(a_n) , where an=∑k=1n{⌊2k−12⌋2}21−k,a_n=\sum_{k=1}^n\left\{\frac{\left\lfloor2^{k-\frac12}\right\rfloor}2\right\}2^{1-k}, converges, and determine its limit as n→∞n\to\infty .

Problem 3

Let pp be a real number and f(x)=xp−x+pf(x)=x^p-x+p . Prove that:

(a) Every root α\alpha of f(x)f(x) satisfies ∣α∣<p1p−1|\alpha|<p^{\frac1{p-1}} ; (b) If pp is a prime number, then f(x)f(x) cannot be written as the product of two non-constant polynomials with integer coefficients.

Problem 4

At each of the given nn points on a circle, either +1+1 or −1-1 is written. The following operation is performed: between any two consecutive numbers on the circle their product is written, and the initial nn numbers are deleted. Suppose that, for any initial arrangement of +1+1 and −1-1 on the circle, after finitely many operations all the numbers on the circle will be equal to +1+1 . Prove that nn is a power of two.

Problem 5

Prove that the perpendiculars, drawn from the midpoints of the edges of the base of a given tetrahedron to the opposite lateral edges, have a common point if and only if the circumcenter of the tetrahedron, the centroid of the base, and the top vertex of the tetrahedron are collinear.

Problem 6

Let x,y,zx,y,z be pairwise coprime positive integers and p≥5p\ge5 and qq be prime numbers which satisfy the following conditions:

(i) 6p6p does not divide q−1q-1 ; (ii) qq divides x2+xy+y2x^2+xy+y^2 ; (iii) qq does not divide x+y−zx+y-z .

Prove that xp+yp≠zpx^p+y^p\ne z^p .