Partition {1,…,1989} into 117 equal-sum 17-element subsets
Prove that the set can be expressed as the disjoint union of subsets () such that:
(i) Each contains 17 elements;
(ii) The sum of all the elements in each is the same.
Topic: Combinatoria, Teoria dei Numeri Metodo: method_casework, Invarianti Abilita: Modellizzazione, Conteggio sistematico, Manipolazione algebrica Area: Combinatoria, Logica e Probabilita, Aritmetica e Teoria dei Numeri Fonte: apri PDF p.1
Partition {1,…,1989} into 117 equal-sum 17-element subsets*
Prove that the set can be expressed as the disjoint union of subsets () such that:
(i) Each contains 17 elements;
(ii) The sum of all the elements in each is the same.
Triangle bisectors, circumcircle points, area relations
In an acute-angled triangle the internal bisector of angle meets the circumcircle of the triangle again at . Points and are defined similarly. Let be the point of intersection of the line with the external bisectors of angles and . Points and are defined similarly. Prove that:
(i) The area of the triangle is twice the area of the hexagon .
(ii) The area of the triangle is at least four times the area of the triangle .
Topic: Geometria piana Metodo: Trigonometria, Simmetria Abilita: Ragionamento geometrico, Manipolazione algebrica, Lettura attenta Area: Geometria Fonte: apri PDF p.1
Triangle bisectors, circumcircle points, area relations
In an acute-angled triangle the internal bisector of angle meets the circumcircle of the triangle again at . Points and are defined similarly. Let be the point of intersection of the line with the external bisectors of angles and . Points and are defined similarly. Prove that:
(i) The area of the triangle is twice the area of the hexagon .
(ii) The area of the triangle is at least four times the area of the triangle .
Point set with equidistance condition implies k < 1/2 + sqrt(2n)
Let and be positive integers and let be a set of points in the plane such that
(i) No three points of are collinear, and
(ii) For any point of there are at least points of equidistant from .
Prove that:
Topic: Combinatoria, Geometria piana Metodo: Doppio conteggio, Estremalità, Disuguaglianze Abilita: Modellizzazione, Stima, Ragionamento geometrico, Manipolazione algebrica Area: Combinatoria, Logica e Probabilita, Geometria Fonte: apri PDF p.1
*Point set with equidistance condition implies k < 1/2 + sqrt(2n) *
Let and be positive integers and let be a set of points in the plane such that
(i) No three points of are collinear, and
(ii) For any point of there are at least points of equidistant from .
Prove that:
Convex quadrilateral with special point yields inequality 1/sqrt(h) >= 1/sqrt(AD)+1/sqrt(BC)
Let be a convex quadrilateral such that the sides , , satisfy . There exists a point inside the quadrilateral at a distance from the line such that and . Show that:
Topic: Geometria piana, Disuguaglianze Metodo: Disuguaglianze, Trigonometria Abilita: Ragionamento geometrico, Manipolazione algebrica, Modellizzazione Area: Geometria, Algebra e Analisi Fonte: apri PDF p.1
*Convex quadrilateral with special point yields inequality 1/sqrt(h) >= 1/sqrt(AD) +1/sqrt(BC) *
Let be a convex quadrilateral such that the sides , , satisfy . There exists a point inside the quadrilateral at a distance from the line such that and . Show that:
For each n, find n consecutive integers none a prime power
Prove that for each positive integer there exist consecutive positive integers none of which is an integral power of a prime number.
Topic: Teoria dei Numeri, Combinatoria Metodo: method_casework, Congruenze, Fattorizzazione Abilita: Modellizzazione, Manipolazione algebrica, Riconoscimento di pattern Area: Aritmetica e Teoria dei Numeri, Combinatoria, Logica e Probabilita Fonte: apri PDF p.1
For each n, find n consecutive integers none at prime power
Prove that for every positive integer there exist consecutive positive integers none of which is an integral power of a prime number.
Permutations of {1,…,2n} with property P outnumber those without
A permutation of the set , where is a positive integer, is said to have property if for at least one in . Show that, for each , there are more permutations with property than without.
Topic: Combinatoria Metodo: Doppio conteggio, Biiezione, Inclusione-esclusione, Conteggio Abilita: Conteggio sistematico, Modellizzazione, Manipolazione algebrica, Ragionamento geometrico Area: Combinatoria, Logica e Probabilita Fonte: apri PDF p.1
Permutations of {1,…,2n} with property P outnumber those without*
A permutation of the set , where is a positive integer, is said to have property if for at least one in . Show that, for each , there are more permutations with property than without.