Find EG/EF in terms of t for tangent-line circle config
Chords AB and CD of a circle intersect at a point E inside the circle. Let M be an interior point of the segment EB. The tangent line at E to the circle through D, E, and M intersects the lines BC and AC at F and G, respectively. If AM AB = t, find EG EF in terms of t.
Topic: Geometria piana Metodo: Sfruttamento della simmetria Abilita: Ragionamento geometrico Area: Geometria Fonte: apri PDF p.1
Find EG/EF in terms of t for tangent-line circle config
Chords AB and CD of a circle intersect at a point E inside the circle. Let M be an interior point of the EB segment. The tangent line at E to the circle through D, E, and M intersects the lines BC and AC at F and G, respectively. If AM AB = t, find EG EF in terms of t.
Smallest k making every coloring of 2n-1 points good
Let n ≥3 and consider a set E of 2n −1 distinct points on a circle. Suppose that exactly k of these points are to be colored black. Such a coloring is “good” if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly n points from E. Find the smallest value of k so that every such coloring of k points of E is good.
Topic: Combinatoria Metodo: Principio dei cassetti Area: Combinatoria, Logica e Probabilita Fonte: apri PDF p.1
Smallest k making every coloring of 2n-1 points good
Let n ≥3 and consider a set E of 2n −1 distinct points on a circle. Suppose that exactly k of these points are to be colored black. Such a coloring is good if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly n points from E. Find the smallest value of k so that every such coloring of k points of E is good.
Determine all n>1 with n^2 dividing 2^n+1
Determine all integers n > 1 such that 2n + 1 n2 is an integer.
31st International Mathematical Olympiad Beijing, China Day II July 13, 1990
Topic: Teoria dei Numeri Metodo: congruenze Area: Aritmetica e Teoria dei Numeri Fonte: apri PDF p.1
Determining at n>1 with n^2 dividing 2^n+1
Determine to the integers n > 1 such that 2n + 1 n2 is an integer.
31st International Mathematical Olympiad Beijing, China Day II July 13, 1990
Construct function on positive rationals with f(xf(y))=f(x)/y
Let Q+ be the set of positive rational numbers. Construct a function f : Q+ → Q+ such that f(xf(y)) = f(x) y for all x, y in Q+.
Topic: successioni Abilita: generalizzazione Area: Algebra e Analisi Fonte: apri PDF p.2
Construct function on positive rationals with f(xf(y))=f(x)/y*
Let Q+ be the set of positive rational numbers. Construct a function f: Q+ → Q+ such that f(xf(y)) = f(x) y for all x, y in Q+.
winning strategies by n0
Given an initial integer n0 > 1, two players, A and B, choose integers n1, n2, n3, … alternately according to the following rules: Knowing n2k, A chooses any integer n2k+1 such that n2k ≤n2k+1 ≤n2 2k. Knowing n2k+1, B chooses any integer n2k+2 such that n2k+1 n2k+2 is a prime raised to a positive integer power. Player A wins the game by choosing the number 1990; player B wins by choosing the number 1. For which n0 does: (a) A have a winning strategy? (b) B have a winning strategy? (c) Neither player have a winning strategy?
Topic: Logica, giochi, strategie, Teoria dei Numeri Metodo: monovarianti Area: Aritmetica e Teoria dei Numeri, Combinatoria, Logica e Probabilita Fonte: apri PDF p.2
winning strategies by n0
Given an initial integer n0 > 1, two players, A and B, choose integers n1, n2, n3, … alternately according to the following rules: Knowing n2k, A chooses any integer n2k+1 such that n2k ≤n2k+1 ≤n2 2k. Knowing n2k+1, B chooses any integer n2k+2 such that n2k+1 n2k+2 is a prime raised to a positive integer power. Player A wins the game by choosing the number 1990; player B wins by choosing the number 1. For which n0 does: (a) A have a winning strategy? (b) B have a winning strategy? (c) Neither player has a winning strategy?
Construct equiangular 1990-gon with square-number side lengths
Prove that there exists a convex 1990-gon with the following two properties: (a) All angles are equal. (b) The lengths of the 1990 sides are the numbers 12, 22, 32, … , 19902 in some order.
Topic: Geometria piana, Combinatoria Metodo: Sfruttamento della simmetria Area: Combinatoria, Logica e Probabilita, Geometria Fonte: apri PDF p.2
Construct an equivalency 1990-gon with square-number side lengths
Prove that there exists a convex 1990-gon with the following two properties: (a) All angles are equal. (b) The lengths of the 1990 sides are the numbers 12, 22, 32… In some order.