MathDB

4

Part of 1967 IMO Shortlist

Problems(10)

IMO LongList 1967, Bulgaria 4

Source: IMO LongList 1967, Bulgaria 4

11/14/2004
Suppose medians mam_a and mbm_b of a triangle are orthogonal. Prove that: a.) Using medians of that triangle it is possible to construct a rectangular triangle. b.) The following inequality: 5(a2+b2−c2)≥8ab,5(a^2+b^2-c^2) \geq 8ab, is valid, where a,ba,b and cc are side length of the given triangle.
geometrygeometric inequalityconstructionmedianTriangleIMO ShortlistIMO Longlist
IMO LongList 1967, Hungary 4

Source: IMO LongList 1967, Hungary 4

12/16/2004
Let k1k_1 and k2k_2 be two circles with centers O1O_1 and O2O_2 and equal radius rr such that O1O2=rO_1O_2 = r. Let AA and BB be two points lying on the circle k1k_1 and being symmetric to each other with respect to the line O1O2O_1O_2. Let PP be an arbitrary point on k2k_2. Prove that PA2+PB2≥2r2.PA^2 + PB^2 \geq 2r^2.
analytic geometrygeometrygeometric inequalitycirclesIMO ShortlistIMO Longlist
Arithmetic progression in equation system

Source: IMO Longlist 1967, Mongolia 4

10/14/2005
In what case does the system of equations x+y+mz=ax+my+z=bmx+y+z=c\begin{matrix} x + y + mz = a \\ x + my + z = b \\ mx + y + z = c \end{matrix} have a solution? Find conditions under which the unique solution of the above system is an arithmetic progression.
linear algebramatrixalgebrasystem of equationsarithmetic sequenceIMO ShortlistIMO Longlist
IMO LongList 1967, Italy 4

Source: IMO LongList 1967, Italy 4

12/16/2004
Find values of the parameter uu for which the expression y=tan⁡(x−u)+tan⁡(x)+tan⁡(x+u)tan⁡(x−u)tan⁡(x)tan⁡(x+u)y = \frac{ \tan(x-u) + \tan(x) + \tan(x+u)}{ \tan(x-u)\tan(x)\tan(x+u)} does not depend on x.x.
parameterizationcalculustrigonometryalgebraTrigonometric IdentitiesIMO Longlist
3n^2 + 3n + 7

Source: IMO Longlist 1967, Poland 4

10/14/2005
Does there exist an integer such that its cube is equal to 3n2+3n+7,3n^2 + 3n + 7, where nn is an integer.
modular arithmeticnumber theoryperfect cubeDiophantine equationIMO ShortlistIMO Longlist
IMO LongList 1967, Romania 4

Source: IMO LongList 1967, Romania 4

12/16/2004
(i) Solve the equation: sin⁡3(x)+sin⁡3(2π3+x)+sin⁡3(4π3+x)+34cos⁡2x=0. \sin^3(x) + \sin^3\left( \frac{2 \pi}{3} + x\right) + \sin^3\left( \frac{4 \pi}{3} + x\right) + \frac{3}{4} \cos {2x} = 0. (ii) Supposing the solutions are in the form of arcs ABAB with one end at the point AA, the beginning of the arcs of the trigonometric circle, and PP a regular polygon inscribed in the circle with one vertex in AA, find: 1) The subsets of arcs having the other end in BB in one of the vertices of the regular dodecagon. 2) Prove that no solution can have the end BB in one of the vertices of polygon PP whose number of sides is prime or having factors other than 2 or 3.
trigonometryalgebraTrigonometric EquationsgeometryIMO ShortlistIMO Longlist
IMO LongList 1967, Socialists Republic Of Czechoslovakia 4

Source: IMO LongList 1967, Socialists Republic Of Czechoslovakia 4

12/16/2004
The square ABCDABCD has to be decomposed into nn triangles (which are not overlapping) and which have all angles acute. Find the smallest integer nn for which there exist a solution of that problem and for such nn construct at least one decomposition. Answer whether it is possible to ask moreover that (at least) one of these triangles has the perimeter less than an arbitrarily given positive number.
geometryperimetersquaredissectiontriangulationIMO ShortlistIMO Longlist
IMO LongList 1967, Sweden 4

Source: IMO LongList 1967, Sweden 4

12/16/2004
A subset SS of the set of integers 0 - 99 is said to have property AA if it is impossible to fill a crossword-puzzle with 2 rows and 2 columns with numbers in SS (0 is written as 00, 1 as 01, and so on). Determine the maximal number of elements in the set SS with the property A.A.
combinatoricsExtremal combinatoricscountingIMO ShortlistIMO Longlist
IMO LongList 1967, The Democratic Republic Of Germany 4

Source: IMO LongList 1967, The Democratic Republic Of Germany 4

12/16/2004
Prove the following statement: If r1r_1 and r2r_2 are real numbers whose quotient is irrational, then any real number xx can be approximated arbitrarily well by the numbers of the form  zk1,k2=k1r1+k2r2\ z_{k_1,k_2} = k_1r_1 + k_2r_2 integers, i.e. for every number xx and every positive real number pp two integers k1k_1 and k2k_2 can be found so that ∣x−(k1r1+k2r2)∣<p|x - (k_1r_1 + k_2r_2)| < p holds.
number theoryapproximationirrational number
IMO LongList 1967, Soviet Union 4

Source: IMO LongList 1967, Soviet Union 4

12/16/2004
In a group of interpreters each one speaks one of several foreign languages, 24 of them speak Japanese, 24 Malaysian, 24 Farsi. Prove that it is possible to select a subgroup in which exactly 12 interpreters speak Japanese, exactly 12 speak Malaysian and exactly 12 speak Farsi.
combinatoricsSet systemsIMO ShortlistIMO Longlist