MathDB

6

Part of 2014 JBMO Shortlist

Problems(3)

AZE JBMO TST

Source: AZE JBMO TST

5/2/2015
Let a,b,ca,b,c be positive real numbers. Prove that ((3a2+1)2+2(1+3b)2)((3b2+1)2+2(1+3c)2)((3c2+1)2+2(1+3a)2)≥483\left((3a^2+1)^2+2\left(1+\frac{3}{b}\right)^2\right)\left((3b^2+1)^2+2\left(1+\frac{3}{c}\right)^2\right)\left((3c^2+1)^2+2\left(1+\frac{3}{a}\right)^2\right)\geq 48^3
inequalities
Nice and interesting JBMO problem

Source: JBMO Shortlist 2014,G6

11/6/2016
Let ABCDABCD be a quadrilateral whose diagonals are not perpendicular and whose sides ABAB and CDCD are not parallel.Let OO be the intersection of its diagonals.Denote with H1H_1 and H2H_2 the orthocenters of triangles AOBAOB and COD,COD, respectively.If MM and NN are the midpoints of the segment lines ABAB and CD,CD, respectively.Prove that the lines H1H2H_1H_2 and MNMN are parallel if and only if AC=BD.AC=BD.
geometry
find 3 consecutive positive integers given 5 conditions

Source: JBMO Shortlist 2014 NT6

4/24/2019
Vukasin, Dimitrije, Dusan, Stefan and Filip asked their teacher to guess three consecutive positive integers, after these true statements: Vukasin: " The sum of the digits of one number is prime number. The sum of the digits of another of the other two is, an even perfect number.(nn is perfect if σ(n)=2n\sigma\left(n\right)=2n). The sum of the digits of the third number equals to the number of it's positive divisors". Dimitrije:"Everyone of those three numbers has at most two digits equal to 11 in their decimal representation". Dusan:"If we add 1111 to exactly one of them, then we have a perfect square of an integer" Stefan:"Everyone of them has exactly one prime divisor less than 1010". Filip:"The three numbers are square free". Professor found the right answer. Which numbers did he mention?
number theoryconsecutivepositive integerssquarefreedecimal representation