Partition 1..1989 into 117 equal-sum 17-element subsets
Prove that the set {1, 2, … , 1989} can be expressed as the disjoint union of subsets Ai (i = 1, 2, … , 117) such that: (i) Each Ai contains 17 elements; (ii) The sum of all the elements in each Ai is the same.
Topic: Combinatoria Metodo: Conteggio combinatorio Area: Combinatoria, Logica e Probabilita Fonte: apri PDF p.1
Partition 1..1989 into 117 equal-sum 17-element subsets
Prove that the set {1, 2, … , 1989} can be expressed as the disjoint union of subsets Ai (i = 1, 2, … , 117) such that: (i) Each Ai contains 17 elements; (ii) The sum of all the elements in each Ai is the same.
Bisector-defined triangle area at least four times ABC
In an acute-angled triangle ABC the internal bisector of angle A meets the circumcircle of the triangle again at A1. Points B1 and C1 are defined similarly. Let A0 be the point of intersection of the line AA1 with the external bisectors of angles B and C. Points B0 and C0 are defined similarly. Prove that: (i) The area of the triangle A0B0C0 is twice the area of the hexagon AC1BA1CB1. (ii) The area of the triangle A0B0C0 is at least four times the area of the triangle ABC.
Topic: Geometria piana, Disuguaglianze Metodo: Disuguaglianze classiche, Sfruttamento della simmetria Abilita: Ragionamento geometrico Area: Algebra e Analisi, Geometria Fonte: apri PDF p.1
Bisector-defined triangle area at least four times ABC
In an acute-angled triangle ABC the internal bisector of angle A meets the circumcircle of the triangle again at A1. Points B1 and C1 are defined similarly. Let A0 be the point of intersection of the line AA1 with the external bisectors of angles B and C. Points B0 and C0 are defined similarly. Prove that: (i) The area of the triangle A0B0C0 is twice the area of the hexagon AC1BA1CB1. (ii) The area of triangle A0B0C0 is at least four times the area of triangle ABC.
Equidistant-points configuration bound k<1/2+sqrt(2n)
Let n and k be positive integers and let S be a set of n points in the plane such that (i) No three points of S are collinear, and (ii) For any point P of S there are at least k points of S equidistant from P. Prove that: k < 1 2 + √ 2n.
30th International Mathematical Olympiad Braunschweig, Germany Day II
Topic: Combinatoria, Geometria piana Metodo: Doppio conteggio Area: Combinatoria, Logica e Probabilita, Geometria Fonte: apri PDF p.1
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Let n and k be positive integers and let S be a set of n points in the plane such that (i) No three points of S are collinear, and (ii) For any point P of S there are at least k points of S equidistant from P. Prove that: k < 1 2 + √ 2n.
30th International Mathematical Olympiad Braunschweig, Germany Day II
Convex quadrilateral distance inequality with interior point P
Let ABCD be a convex quadrilateral such that the sides AB, AD, BC satisfy AB = AD + BC. There exists a point P inside the quadrilateral at a distance h from the line CD such that AP = h + AD and BP = h + BC. Show that: 1 √ h ≥ 1 √ AD + 1 √ BC .
Topic: Geometria piana, Disuguaglianze Metodo: Disuguaglianze classiche Area: Algebra e Analisi, Geometria Fonte: apri PDF p.2
Convex quadrilateral distance inequality with interior point P*
Let ABCD be a convex quadrilateral such that the sides AB, AD, BC satisfy AB = AD + BC. There exists a point P inside the quadrilateral at a distance h from the line CD such that AP = h + AD and BP = h + BC. Show that: 1 √ h ≥ 1 √ AD + 1 √ BC .
n consecutive integers none a prime power
Prove that for each positive integer n there exist n consecutive positive integers none of which is an integral power of a prime number.
Topic: Teoria dei Numeri Metodo: congruenze Area: Aritmetica e Teoria dei Numeri Fonte: apri PDF p.2
*n consecutive integers none at prime power *
Prove that for every positive integer n there exist n consecutive positive integers none of which is an integral power of a prime number.
More permutations of 1..2n with property P than without
A permutation (x1, x2, … , xm) of the set {1, 2, … , 2n}, where n is a positive integer, is said to have property P if |xi −xi+1| = n for at least one i in {1, 2, … , 2n −1}. Show that, for each n, there are more permutations with property P than without.
Topic: Combinatoria Metodo: corrispondenza, Conteggio combinatorio Area: Combinatoria, Logica e Probabilita Fonte: apri PDF p.2
More permutations of 1..2n with property P than without
A permutation (x1, x2, … , xm) of the set {1, 2, … , 2n}, where n is a positive integer, is said to have property P if xxi −xi+1 = n for at least one i in {1, 2, … , 2n −1}. Show that, for each n, there are more permutations with property P than without.