MathDB

5

Part of 2012 Dutch IMO TST

Problems(2)

collinear wanted, circumcircle related

Source: Dutch IMO TST1 2012 p5

8/6/2019
Let Γ\Gamma be the circumcircle of the acute triangle ABCABC. The angle bisector of angle ABCABC intersects ACAC in the point B1B_1 and the short arc ACAC of Γ\Gamma in the point PP. The line through B1B_1 perpendicular to BCBC intersects the short arc BCBC of Γ\Gamma in KK. The line through BB perpendicular to AKAK intersects ACAC in LL. Prove that K,LK, L and PP lie on a line.
geometryangle bisectorcircumcircle
f(x + xy + f(y))=(f(x) + 1/2)(f(y) +1/2)

Source: Dutch IMO TST2 2012 p5

1/10/2020
Find all functions f:R→Rf : R \to R satisfying f(x+xy+f(y))=(f(x)+12)(f(y)+12)f(x + xy + f(y))=(f(x) + \frac12)(f(y) + \frac12 ) for all x,y∈Rx, y \in R.
functional equationalgebra