olymmath

BG-MO 1991

Bulgarian Mathematical Olympiad 1991.

Other problems

Problem 1

Let MM be a point on the altitude CDCD of an acute-angled triangle ABCABC , and KK and LL the orthogonal projections of MM on ACAC and BCBC . Suppose that the incenter and circumcenter of the triangle lie on the segment KLKL .

(a) Prove that CD=R+rCD=R+r , where RR and rr are the circumradius and inradius, respectively. (b) Find the minimum value of the ratio CM:CDCM:CD .

Problem 2

Let KK be a cube with edge nn , where n>2n>2 is an even integer. Cube KK is divided into n3n^3 unit cubes. We call any set of n2n^2 unit cubes lying on the same horizontal or vertical level a layer. We dispose of n34\frac{n^3}4 colors, in each of which we paint exactly 44 unit cubes. Prove that we can always select nn unit cubes of distinct colors, no two of which lie on the same layer.

Problem 3

Prove that for every prime number p≥5p\ge5 ,

(a) p3p^3 divides (2pp)−2\binom{2p}p-2 ; (b) p3p^3 divides (kpp)−k\binom{kp}p-k for every natural number kk .

Problem 4

Let f(x)f(x) be a polynomial of degree nn with real coefficients, having nn (not necessarily distinct) real roots. Prove that for all real xx , f(x)f′′(x)≤f′(x)2.f(x)f''(x)\le f'(x)^2.

Problem 5

On a unit circle with center OO , ABAB is an arc with the central angle α<90∘\alpha<90^\circ . Point HH is the foot of the perpendicular from AA to OBOB , TT is a point on arc ABAB , and ll is the tangent to the circle at TT . The line ll and the angle AHBAHB form a triangle Δ\Delta .

(a) Prove that the area of Δ\Delta is minimal when TT is the midpoint of arc ABAB . (b) Prove that if SαS_\alpha is the minimal area of Δ\Delta then the function Sαα\frac{S_\alpha}\alpha has a limit when α→0\alpha\to0 and find this limit.

Problem 6

White and black checkers are put on the squares of an n×nn\times n chessboard (n≥2)(n\ge2) according to the following rule. Initially, a black checker is put on an arbitrary square. In every consequent step, a white checker is put on a free square, whereby all checkers on the squares neighboring by side are replaced by checkers of the opposite colors. This process is continued until there is a checker on every square. Prove that in the final configuration there is at least one black checker.