MathDB

2

Part of 2014 Belarus Team Selection Test

Problems(6)

locus of midpoints wanted, circle related

Source: 2014 Belarus TST 1.2

12/29/2020
Given a triangle ABCABC. Let SS be the circle passing through CC, centered at AA. Let XX be a variable point on SS and let KK be the midpoint of the segment CXCX . Find the locus of the midpoints of BKBK, when XX moves along SS.
(I. Gorodnin)
geometrymidpointLocus
(a + b + c)(ab + bc+ca) + 3abc>= 4(ab + bc + ca) if ab+bc+ca>= a+b+c

Source: 2014 Belarus TST 2.2

12/30/2020
Given positive real numbers a,b,ca,b,c with ab+bc+ca≥a+b+cab+bc+ca\ge a+b+c , prove that (a+b+c)(ab+bc+ca)+3abc≥4(ab+bc+ca).(a + b + c)(ab + bc+ca) + 3abc \ge 4(ab + bc + ca).
(I. Gorodnin)
algebrainequalities
3-var ineq, a+b+c=1

Source: Belarus 2014

2/22/2020
Let a,b,ca,b,c be positive real numbers such that a+b+c=1a+b+c=1. Prove that a2(b+c)3+b2(c+a)3+c2(a+b)3≥98\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}\geq \frac98
InequalityalgebraBelarus
(x + y + z)xyz= -a^2 if x^2-1/y = y^2 -1/z = z^2 -1/x=a

Source: 2014 Belarus TST 3.2

12/30/2020
Let x,y,zx,y,z be pairwise distinct real numbers such that x2−1/y=y2−1/z=z2−1/xx^2-1/y = y^2 -1/z = z^2 -1/x. Given z2−1/x=az^2 -1/x = a, prove that (x+y+z)xyz=−a2(x + y + z)xyz= -a^2.
(I. Voronovich)
algebrasystem of equations
a_n=a_{a_{n-1}}+a_{a_{n+1}}, of positive integers

Source: 2014 Belarus TST 5.2

12/29/2020
Find all sequences (an)(a_n) of positive integers satisfying the equality an=aan−1+aan+1a_n=a_{a_{n-1}}+a_{a_{n+1}} a) for all n≥2n\ge 2 b) for all n≥3n \ge 3
(I. Gorodnin)
Sequencerecurrence relationalgebranumber theory with sequences
{n\sqrt6 } > 1/n and {n\sqrt6 }> / 1/ (n-1/(5n)) , for odd n

Source: 2014 Belarus TST 8.2

12/29/2020
Prove that for all even positive integers nn the following inequality holds a) {n6}>1n\{n\sqrt6\} > \frac{1}{n} b){n6}>1n−1/(5n) \{n\sqrt6\}> \frac{1}{n-1/(5n)}
(I. Voronovich)
algebrainequalities