Maths and Science discussion forum

ARML local is online

@ChessManIsMe http://www.seedasdan.org/en/arml-en/

I see, so it's basically ARML offering an international contest. Interesting, I didn't know that, thx

Let PA=a, PB=b, PC=c, by cosine formula a^2+ab+b^2=9, b^2+bc+c^2=25, c^2+ca+a^2=49. sqrt(3)/4*(ab+bc+ca)=[ABC]=15sqrt(3)/4, so ab+bc+ca=15. And, 2a^2+2b^2+2c^2+ab+bc+ca=83, so a^2+b^2+c^2+(ab+bc+ca)/2=83/2, so a^2+b^2+c^2+2(ab+bc+ca)=83/2+3/2*15=64, so a+b+c=8

If f(x) is a one-to-one function such that f(f(x))+f(f(y))=f((x+y)/2), what is f(x)? (Source: Me)

this is elementary functional equation

f(x)=c+0.5x, where c is a constant, right? Idk, I’m bad at functional equations but correct me if I’m wrong:

Since f is one to one, take the inverse of both sides. f(x)+f(y)=x/2+y/2

Let f(x)-x/2=g(x). g(x) must equal a constant because g(x)=g(y) for all x, y.

f(x)=c+(x/2)=c+0.5x. Nice problem!

sorry guys, i made a typo, to make it actually ELEMENTARY I need to make a change: (f(f(x))+f(f(y)))/2=f((x+y)/2). This one is super easy

and no, f inverse of (f(f(x)+f(f(y)) might not be f(x)+f(y)

DAILY PROBLEM:

 Given the side lengths of triangle ABC, find the area of its orthic triangle.

haha I see, I read the question wrong lol

(f(f(x))+f(f(y)))/2=f((x+y)/2)

Set x=y because the equation must be true for all x and y,

f(f(x))=f(x).

f inv(f(f(x)))=f inv(f(x))

f(x)=x.

Therefore f(x)=x, right?

yeah, it is really easy

but your solution is not complete

the last step is to verify f(x)=x always works, but that is also very easy

log4 is 0.602059991

it is not, it is approximately

I'm in 8th grade so I still have to learn these.

so yeah