Find all functions $f:Q \rightarrow Q$ such that
\[f(1)=2\]
and
\[f(xy)=f(x)f(y)−f(x+y)+1\]
Search found 98 matches
- Sun Mar 05, 2017 6:42 pm
- Forum: Junior Level
- Topic: Don't Intersect Please :D
- Replies: 7
- Views: 5396
Re: Don't Intersect Please :D
Simple polygon will also do the trick. Isn't? If we put a super big rubber-band, it will result us a simple polygon.
- Thu Mar 02, 2017 8:42 pm
- Forum: Social Lounge
- Topic: A question
- Replies: 1
- Views: 2431
Re: A question
No. ,
- Thu Mar 02, 2017 8:08 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/4
- Replies: 8
- Views: 6178
Re: IGO 2016 Elementary/4
I more solution Case 1:$AC>AB$ By a well know lemma we know that $BLKC$ is cyclic. Now denote, $\angle LBK=\angle LCK=\angle LCB=\angle LKB=a$. And we have that $KC=KB$. So, $\angle KBC=2a$. So, we have $a+2a+a+a=90^o \Rightarrow a=18^o$. That means $2a=\angle C=18^o\times 2=36^o$. So, $\angle B= 90...
- Wed Mar 01, 2017 9:44 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/4
- Replies: 8
- Views: 6178
Re: IGO 2016 Elementary/4
DONE. Didn't use any angle chasing as I saidahmedittihad wrote:Why couldn't you? Thamim
- Wed Mar 01, 2017 7:20 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/4
- Replies: 8
- Views: 6178
Re: IGO 2016 Elementary/4
It is a very nasty problem. By the way, thanks me if you read the solution. It was a hard time to $LateX$ all of them. It might look big, but, it is not very hard. Case 1: $AC > AB$. Claim 1: In $\triangle ABC$, $\angle LBK=\angle LKB$. Proof: Let $X$ be the midpoint of $KB$. We know that, $LX \perp...
- Tue Feb 28, 2017 7:19 pm
- Forum: Junior Level
- Topic: Don't Intersect Please :D
- Replies: 7
- Views: 5396
Don't Intersect Please :D
Given $2n$ points in the plane with no three collinear, show that it is possible to pair them up in such a way that the $n$ line segments joining paired points do not intersect.
- Mon Feb 27, 2017 9:20 pm
- Forum: Junior Level
- Topic: Czech and Slovak Republics 1997
- Replies: 4
- Views: 59747
Czech and Slovak Republics 1997
Each side and diagonal of a regular $2n+1$-gon ($2n+1 \ge 3$) is colored blue or green. A move consists of choosing a vertex and switching the color of each segment incident to that vertex (from blue to green or vice versa). Prove that regardless of the initial coloring, it is possible to make the n...
- Mon Feb 27, 2017 12:21 am
- Forum: Social Lounge
- Topic: ক্যাম্প
- Replies: 4
- Views: 3680
Re: ক্যাম্প
$0=0$. Right?
- Sun Feb 26, 2017 11:12 pm
- Forum: Social Lounge
- Topic: ক্যাম্প
- Replies: 4
- Views: 3680
Re: ক্যাম্প
না। ।