MathDB

3

Part of 2009 Belarus Team Selection Test

Problems(4)

sum cosA + cosB wanted, 2 angle bisectors and an altitude related

Source: Belarus TST 2009 1.3

6/13/2020
Points T,P,HT,P,H lie on the side BC,AC,ABBC,AC,AB respectively of triangle ABCABC, so that BPBP and ATAT are angle bisectors and CHCH is an altitude of ABCABC. Given that the midpoint of CHCH belongs to the segment PT,PT, find the value of cos⁡A+cos⁡B\cos A + \cos B
I. Voronovich
geometryangle bisectoraltitudetrigonometry
parallel in a tangential trapezoid with 2 right angles

Source: Belarus TST 2009 2.3

6/13/2020
Given trapezoid ABCDABCD (AD∥BCAD\parallel BC) with AD⊥ABAD \perp AB and T=AC∩BDT=AC\cap BD. A circle centered at point OO is inscribed in the trapezoid and touches the side CDCD at point QQ. Let PP be the intersection point (different from QQ) of the side CDCD and the circle passing through T,QT,Q and OO. Prove that TP∥ADTP \parallel AD.
I. Voronovich
geometrytrapezoidright angleparalleltangential quadrilateraltangential
x + f(y) =a(y + f(x)) , parameter functional

Source: 2009 Belarus TST 5.3

11/8/2020
Find all real numbers aa for which there exists a function f:R→Rf: R \to R asuch that x+f(y)=a(y+f(x))x + f(y) =a(y + f(x)) for all real numbers x,y∈Rx,y\in R.
I.Voronovich
functionalalgebrafunctional equation
2 different functionals ff(f(n))=f(n+1) - f(n), f(f(n))=f(n+2) - f(n)

Source: 2009 Belarus TST 8.3

11/8/2020
a) Does there exist a function f:N→Nf: N \to N such that f(f(n))=f(n+1)−f(n)f(f(n))=f(n+1) - f(n) for all n∈Nn \in N? b) Does there exist a function f:N→Nf: N \to N such that f(f(n))=f(n+2)−f(n)f(f(n))=f(n+2) - f(n) for all n∈Nn \in N?
I. Voronovich
algebrafunctionalfunctional equation