MathDB

2013 Czech-Polish-Slovak Match

Part of Czech-Polish-Slovak Match

Subcontests

(3)
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x^2-rx and x^3-rx are rationals implies x is rational

For each rational number rr consider the statement: If xx is a real number such that x2−rxx^2-rx and x3−rxx^3-rx are both rational, then xx is also rational.
(a) Prove the claim for r≥43r \ge \frac43 and r≤0r \le 0. (b) Let p,qp,q be different odd primes such that 3p<4q3p <4q. Prove that the claim for r=pqr=\frac{p}q does not hold.

circumcircle passes through the midpoint

Let ABC{ABC} be a triangle inscribed in a circle. Point P{P} is the center of the arc BAC{BAC}. The circle with the diameter CP{CP} intersects the angle bisector of angle ∠BAC{\angle BAC} at points K,L{K, L} (∣AK∣<∣AL∣){(|AK| <|AL|)}. Point M{M} is the reflection of L{L} with respect to line BC{BC}. Prove that the circumcircle of the triangle BKM{BKM} passes through the center of the segment BC{BC} .
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x^n + 1/x^n - 2 &gt; n^2(x + 1/x -2)

Prove that for every real number x>0x>0 and each integer n>0n>0 we have xn+1xn−2≥n2(x+1x−2)x^n+\frac1{x^n}-2 \ge n^2\left(x+\frac1x-2\right)

triangular cells of an equilateral triangle are infected

Triangular grid divides an equilateral triangle with sides of length nn into n2n^2 triangular cells as shown in figure for n=12n=12. Some cells are infected. A cell that is not yet infected, ia infected when it shares adjacent sides with at least two already infected cells. Specify for n=12n=12, the least number of infected cells at the start in which it is possible that over time they will infected all the cells of the original triangle. [asy] unitsize(0.25cm); path p=polygon(3); for(int m=0; m<=11;++m){ for(int n=0 ; n<= 11-m; ++n){ draw(shift((n+0.5*m)*sqrt(3),1.5*m)*p); } } [/asy]
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