MathDB

1

Part of 2015 Saudi Arabia IMO TST

Problems(4)

f (x/y) = f(x) + f(y) - f(x)f(y)

Source: 2015 Saudi Arabia IMO TST I p1

7/24/2020
Find all functions f:R>0→Rf : R_{>0} \to R such that f(xy)=f(x)+f(y)−f(x)f(y)f \left(\frac{x}{y}\right) = f(x) + f(y) - f(x)f(y) for all x,y∈R>0x, y \in R_{>0}. Here, R>0R_{>0} denotes the set of all positive real numbers.
Nguyễn Duy Thái Sơn
algebrafunctionalfunctional equation
if GL bisects HP then P is the incenter of ABC

Source: 2015 Saudi Arabia IMO TST II p1

7/24/2020
Let ABCABC be an acute-angled triangle inscribed in the circle (O)(O), HH the foot of the altitude of ABCABC at AA and PP a point inside ABCABC lying on the bisector of ∠BAC\angle BAC. The circle of diameter APAP cuts (O)(O) again at GG. Let LL be the projection of PP on AHAH. Prove that if GLGL bisects HPHP then PP is the incenter of the triangle ABCABC.
Lê Phúc Lữ
geometryincenter
we can select from a_1, a_2,..., a_k some numbers so that sum of these is S

Source: 2015 Saudi Arabia IMO TST III p1

7/24/2020
Let SS be a positive integer divisible by all the integers 1,2,...,20151, 2,...,2015 and a1,a2,...,aka_1, a_2,..., a_k numbers in {1,2,...,2015}\{1, 2,...,2015\} such that 2S≤a1+a2+...+ak2S \le a_1 + a_2 + ... + a_k. Prove that we can select from a1,a2,...,aka_1, a_2,..., a_k some numbers so that the sum of these selected numbers is equal to SS.
Lê Anh Vinh
combinatoricsSum
gcd(c^2 + d^2, a^2 + b^2) > 1. when ac+bd is divisible by a^2 +b^2

Source: 2015 Saudi Arabia IMO TST IV p1

7/24/2020
Let a,b,c,da, b,c,d be positive integers such that ac+bdac+bd is divisible by a2+b2a^2 +b^2. Prove that gcd(c2+d2,a2+b2)>1gcd(c^2 + d^2, a^2 + b^2) > 1.
Trần Nam Dũng
number theorydividesdivisibleGCD