Chessboard triangle black-white area difference function f(m,n)
In the plane the points with integer coordinates are the vertices of unit squares. The squares are colored alternately black and white (as on a chessboard). For any pair of positive integers m and n, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths m and n, lie along edges of the squares. Let S1 be the total area of the black part of the triangle and S2 be the total area of the white part. Let f(m, n) = |S1 −S2|. (a) Calculate f(m, n) for all positive integers m and n which are either both even or both odd. (b) Prove that f(m, n) ≤1 2 max{m, n} for all m and n. (c) Show that there is no constant C such that f(m, n) < C for all m and n.
Topic: Combinatoria, Geometria piana Metodo: Analisi per casi, parita Area: Combinatoria, Logica e Probabilita, Geometria Fonte: apri PDF p.1
Chessboard triangle black-white area difference function f(m,n)
In the plane the points with integers are the vertices of unit squares. The squares are alternately colored black and white (as on a chessboard). For any pair of positive integers m and n, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths m and n, lie along edges of the squares. Let S1 be the total area of the black part of the triangle and S2 be the total area of the white part. Let f, m, n = S1 − S2 be the same. (a) Calculate f ((m, n) for all positive integers m and n which are either even or both odd. (b) Prove that f (m, n) ≤1 2 max{m, n} for all m and n. (c) Show that there is no constant C such that f ((m, n) < C for all m and n.
Prove AU=TB+TC with perpendicular-bisector construction
The angle at A is the smallest angle of triangle ABC. The points B and C divide the circumcircle of the triangle into two arcs. Let U be an interior point of the arc between B and C which does not contain A. The perpendicular bisectors of AB and AC meet the line AU at V and W, respectively. The lines BV and CW meet at T. Show that AU = TB + TC.
Topic: Geometria piana Metodo: Tecniche trigonometriche Abilita: Ragionamento geometrico Area: Geometria Fonte: apri PDF p.1
Prove AU=TB+TC with perpendicular-bisector construction
The angle at A is the smallest angle of triangle ABC. The points B and C divide the circumcircle of the triangle into two arcs. Let U be an interior point of the arc between B and C which does not contain A. The perpendicular bisectors of AB and AC meet the line AU at V and W, respectively. The lines BV and CW meet at T. Show that AU is TB + TC.
Permutation making weighted sum small under coefficient bounds
Let x1, x2, … , xn be real numbers satisfying the conditions |x1 + x2 + · · + xn| = 1 and |xi| ≤n + 1 2 i = 1, 2, … , n. Show that there exists a permutation y1, y2, … , yn of x1, x2, … , xn such that |y1 + 2y2 + · · + nyn| ≤n + 1 2 .
38th International Mathematical Olympiad Mar del Plata, Argentina Day II July 25, 1997
Topic: Algebra Metodo: Principio di estremalita Area: Algebra e Analisi Fonte: apri PDF p.1
Permutation making weighted sum small under coefficient bounds
Let x1, x2, … , xn be real numbers satisfying the conditions |x1 + x2 + · · + xn| = 1 and |xi| ≤n + 1 2 i = 1, 2, … , n. Show that there exists a permutation y1, y2, … , yn of x1, x2, … , xn such that |y1 + 2y2 + · · + nyn| ≤n + 1 2 .
38th International Mathematical Olympiad Mar del Plata, Argentina Day II July 25, 1997
none for n=1997, exist for infinitely many n
An n × n matrix whose entries come from the set S = {1, 2, … , 2n −1} is called a silver matrix if, for each i = 1, 2, … , n, the ith row and the ith column together contain all elements of S. Show that (a) there is no silver matrix for n = 1997; (b) silver matrices exist for infinitely many values of n.
Topic: Combinatoria Metodo: Analisi per casi, parita Area: Combinatoria, Logica e Probabilita Fonte: apri PDF p.2
none for n=1997, exist for infinitely many n
An n × n matrix whose entries come from the set S = {1, 2, … , 2n −1} is called a silver matrix if, for each i = 1, 2, … , n, the ith row and the ith column together contain all elements of S. Show that (a) there is no silver matrix for n = 1997; (b) silver matrices exist for infinitely many values of n.
Find all integer pairs with a^(b^2)=b^a
Find all pairs (a, b) of integers a, b ≥1 that satisfy the equation ab2 = ba.
Topic: Teoria dei Numeri Metodo: Analisi per casi, congruenze Area: Aritmetica e Teoria dei Numeri Fonte: apri PDF p.2
Find the integer pairs with a^(b^2)=b^a
Find all pairs (a, b) of integers a, b ≥1 that satisfy the equation ab2 = ba.
Bound f(2^n), partitions of n into powers of 2
For each positive integer n , let f(n) denote the number of ways of representing n as a sum of powers of 2 with nonnegative integer exponents. Representations which differ only in the ordering of their summands are considered to be the same. For instance, f(4) = 4, because the number 4 can be represented in the following four ways: 4; 2 + 2; 2 + 1 + 1; 1 + 1 + 1 + 1. Prove that, for any integer n ≥3, 2n2/4 < f(2n) < 2n2/2.
Topic: Combinatoria, Teoria dei Numeri Metodo: Induzione, Ricorsione Area: Aritmetica e Teoria dei Numeri, Combinatoria, Logica e Probabilita Fonte: apri PDF p.2
*Bound f(2^n), partitions of n into powers of 2 *
For each positive integer n , let f(n) denotes the number of ways of representing n as a sum of powers of 2 with nonnegative integer exponents. Representations which differ only in the ordering of their summands are considered to be the same. For example, f(4) = 4, because the number 4 can be represented in the following four ways: 4; 2 + 2; 2 + 1 + 1; 1 + 1 + 1 + 1. Prove that, for any integer n ≥3, 2n2/4 < f(2n) < 2n2/2.