olymmath

OMM 2014

Mexican Mathematical Olympiad 2014.

Other problems

Problem 1

Each of the integers from 1 to 4027 has been colored either green or red. Changing the color of a number is making it red if it was green and making it green if it was red. Two positive integers mm and nn are said to be cuates if either mn\frac{m}{n} or nm\frac{n}{m} is a prime number. A step consists in choosing two numbers that are cuates and changing the color of each of them. Show it is possible to apply a sequence of steps such that every integer from 1 to 2014 is green.

Problem 2

A positive integer aa is said to reduce to a positive integer bb if when dividing aa by its units digits the result is bb . For example, 2015 reduces to 20155=403\frac{2015}{5} = 403 . Find all the positive integers that become 1 after some amount of reductions. For example, 12 is one such number because 12 reduces to 6 and 6 reduces to 1.

Problem 3

Let Γ1\Gamma_1 be a circle and PP a point outside of Γ1\Gamma_1 . The tangents from PP to Γ1\Gamma_1 touch the circle at AA and BB . Let MM be the midpoint of PAPA and Γ2\Gamma_2 the circle through PP , AA and BB . Line BMBM cuts Γ2\Gamma_2 at CC , line CACA cuts Γ1\Gamma_1 at DD , segment DBDB cuts Γ2\Gamma_2 at EE and line PEPE cuts Γ1\Gamma_1 at FF , with EE in segment PFPF . Prove lines AFAF , BPBP , and CECE are concurrent.

Problem 4

Problem 4

Let ABCDABCD be a rectangle with diagonals ACAC and BDBD . Let EE be the intersection of the bisector of ∠CAD\angle CAD with segment CDCD , FF on CDCD such that EE is midpoint of DFDF , and GG on BCBC such that BG=ACBG = AC (with CC between BB and GG ). Prove that the circumference through DD , FF and GG is tangent to BGBG .

Problem 5

Let a,b,ca, b, c be positive reals such that a+b+c=3a + b + c = 3 . Prove:

a2a+bc3+b2b+ca3+c2c+ab3≥32\frac{a^2}{a + \sqrt[3]{bc}} + \frac{b^2}{b + \sqrt[3]{ca}} + \frac{c^2}{c + \sqrt[3]{ab}} \geq \frac{3}{2}

And determine when equality holds.

Problem 6

Let d(n)d(n) be the number of positive divisors of a positive integer nn (including 11 and nn ). Find all values of nn such that n+d(n)=d(n)2n + d(n) = d(n)^2 .