olymmath

CZ-SK-MO 1999

Czech-Slovak Mathematical Olympiad 1999.

Other problems

Problem 1

We are allowed to put several brackets in the expression 29:28:27:26:...:17:1615:14:13:12:...:3:2\frac{29 : 28 : 27 : 26 :... : 17 : 16}{15 : 14 : 13 : 12 : ... : 3 : 2} always in the same places below each other. (a) Find the smallest possible integer value we can obtain in that way. (b) Find all possible integer values that can be obtained.

Remark: in this problem, (29:28):27:...:16(15:14):13:...:2,\frac{(29 : 28) : 27 : ... : 16}{(15 : 14) : 13 : ... : 2}, is valid position of parenthesis, on the other hand (29:28):27:...:1615:(14:13):...:2\frac{(29 : 28) : 27 : ... : 16}{15 : (14 : 13) : ... : 2} is forbidden.

Problem 2

In a tetrahedron ABCD,EABCD, E and FF are the midpoints of the medians from AA and DD . Find the ratio of the volumes of tetrahedra BCEFBCEF and ABCDABCD .

Note: Median in a tetrahedron connects a vertex and the centroid of the opposite side.

Problem 3

Show that there exists a triangle ABCABC such that a≠ba \ne b and a+ta=b+tba+t_a = b+t_b , where ta,tbt_a,t_b are the medians corresponding to a,ba,b , respectively. Also prove that there exists a number kk such that every such triangle satisfies a+ta=b+tb=k(a+b)a+t_a = b+t_b = k(a+b) . Finally, find all possible ratios a:ba : b in such triangles.

Problem 4

In a certain language there are only two letters, AA and BB . We know that (i) There are no words of length 11 , and the only words of length 22 are ABAB and BBBB . (ii) A segment of length n>2n > 2 is a word if and only if it can be obtained from a word of length less than nn by replacing each letter BB by some (not necessarily the same) word. Prove that the number of words of length nn is equal to 2n+2⋅(−1)n3\frac{2^n +2\cdot (-1)^n}{3}

Problem 5

Given an acute angle APXAPX in the plane, construct a square ABCDABCD such that PP lies on the side BCBC and ray PXPX meets CDCD in a point QQ such that APAP bisects the angle BAQBAQ .

Problem 6

Find all pairs of real numbers a,ba,b for which the system of equations {x+yx2+y2=ax3+y3x2+y2=b\begin{cases} \dfrac{x+y}{x^2 +y^2} = a \dfrac{x^3 +y^3}{x^2 +y^2} = b \end{cases} has a real solution.