olymmath

CAPS 1996

Czech-Austrian-Polish-Slovak Match 1996.

Other problems

Problem 1

Show that an integer p>3p > 3 is a prime if and only if for every two nonzero integers a,ba,b exactly one of the numbers N1=a+b−6ab+p−16N_1 = a+b-6ab+\frac{p-1}{6} , N2=a+b+6ab+p−16N_2 = a+b+6ab+\frac{p-1}{6} is a nonzero integer.

Problem 2

Let ⋆ be a binary operation on a nonempty set MM . That is, every pair (a,b)∈M(a,b) \in M is assigned an element aa ⋆ bb in MM . Suppose that ⋆ has the additional property that (a(a ⋆ b)b) ⋆ b=ab= a and aa ⋆ (a(a ⋆ b)=bb)= b for all a,b∈Ma,b \in M . (a) Show that aa ⋆ b=bb = b ⋆ aa for all a,b∈Ma,b \in M . (b) On which finite sets MM does such a binary operation exist?

Problem 3

The base of a regular quadrilateral pyramid π\pi is a square with side length 2a2a and its lateral edge has length a 17\sqrt{17} . Let MM be a point inside the pyramid. Consider the five pyramids which are similar to π\pi , whose top vertex is at MM and whose bases lie in the planes of the faces of π\pi . Show that the sum of the surface areas of these five pyramids is greater or equal to one fifth the surface of π\pi , and find for which MM equality holds.

Problem 4

Decide whether there exists a function f:Z→Zf : Z \rightarrow Z such that for each k=0,1,...,1996k =0,1, ...,1996 and for any integer mm the equation f(x)+kx=mf (x)+kx = m has at least one integral solution xx .

Problem 5

Two sets of intervals A,BA ,B on the line are given. The set AA contains 2m−12m-1 intervals, every two of which have an interior point in common. Moreover, every interval from AA contains at least two disjoint intervals from BB . Show that there exists an interval in BB which belongs to at least mm intervals from AA .

Problem 6

The points EE and DD lie in the interior of sides ACAC and BCBC , respectively, of a triangle ABCABC . Let FF be the intersection of the lines ADAD and BEBE .Show that the area of the traingles ABCABC and ABFABF satisfies:

SABCSABF=∣AC∣∣AE∣+∣BC∣∣BD∣−1\frac{S_{ABC}}{S_{ABF}} = \frac{\mid{AC}\mid}{\mid{AE} \mid} + \frac{\mid{BC}\mid}{\mid{BD}\mid} - 1 .