Search found 176 matches
- Tue Feb 14, 2017 6:52 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 5
- Replies: 15
- Views: 11403
Re: BDMO 2017 National round Secondary 5
The length of arc $BC$ is $12$, so the circumference of the circle is $72$. Now, we apply cartesian coordinate. Let $AB=2k$. Take $A$ as the origin and $AB$ as the $x$-axis. Now, define points. 1. $T$: touchpoint of the small circle and the circle with center $A$. 2. $M$: midpoint of $AB$. It has co...
- Tue Feb 14, 2017 6:34 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 National Round Secondary 10
- Replies: 5
- Views: 4739
Re: BdMO 2017 National Round Secondary 10
This was also the problem 10 for higher secondary. Let $1\le u\le p-1$ be an integer. We claim that there is exactly one number $k$ so that $u|pk+1$ and $1\le k\le u$. Now, consider the solutions of the linear diophantine equation $ux-py=1$. Since $(u,p)=1$, there exists infinitely many solutions. L...
- Thu Feb 02, 2017 7:54 pm
- Forum: Algebra
- Topic: A beautiful FE
- Replies: 2
- Views: 2477
Re: A beautiful FE
Let $P(x,y)$ denote the FE, and also assume that $c=f(1)$. Now, $P(1,1)\Rightarrow f(c)=\dfrac{c^2}{2}+1$ $P(c,x)\Rightarrow f(cf(x))+f(f(x))=(\dfrac{c^2}{2}+1)f(x)+2$ $P(f(x),1)\Rightarrow f(c(f(x))+\dfrac{c^2}{2}+1=cf(f(x))+2$ From which we get $f(f(x))=\dfrac{c^2/2+1}{c+1}f(x)+\dfrac{c^2/2+1}{c+1...
- Thu Feb 02, 2017 7:38 pm
- Forum: Algebra
- Topic: A beautiful FE
- Replies: 2
- Views: 2477
A beautiful FE
Find all continuous functions $f:\mathbb{R}^{\ge 0}\rightarrow \mathbb{R}^{\ge 0}$ so that for any $x,y$ in the domain, we have:
\[f(xf(y))+f(f(y))=f(x)f(y)+2\]
\[f(xf(y))+f(f(y))=f(x)f(y)+2\]
- Thu Feb 02, 2017 2:12 pm
- Forum: Junior Level
- Topic: BDMO National Junior 2016/6
- Replies: 9
- Views: 10661
Re: BDMO National Junior 2016/6
All you did was drop the word WLOG. You can't just assume that.dshasan wrote: Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
The correct way to write that is as follows: let $\{3^{2w},3^{3x},3^{5y}\}=\{3^{\alpha},3^{\beta},3^{\gamma}\}$.
where $\alpha < \beta < \gamma$.
- Thu Feb 02, 2017 1:57 pm
- Forum: Algebra
- Topic: 2009 IMO SL A3
- Replies: 2
- Views: 3003
Re: 2009 IMO SL A3
Let $\triangle (a,b,c)$ denote that lengths $a,b,c$ form a valid triangle. Let $f(2)=t+1$, then $P(2,b)\Rightarrow \left|f(b)-f(b+t)\right|= 1$ (we can discard $0$ since $f$ is bijective). If $f(b+t)=f(b)-1$ then by induction $f(b+st)=f(b)-s$ for all $s\in\mathbb N$ that gives a contradiction for f...
- Wed Feb 01, 2017 6:49 pm
- Forum: Number Theory
- Topic: $2008$ ISL N$5$
- Replies: 2
- Views: 2683
Re: $2008$ ISL N$5$
Beautiful problem!!! Lemma 1: $f(2)=2, f(3)=3^2$. Proof: $d(f(2))=2$, so $f(2)=p$ where $p$ is a prime. Also, $d(f(3))=3$, so $f(3)=q^2$ where $q$ is a prime. Now, $f(6)=f(2\times 3)|3^5p$ $f(6)=f(3\times 2)|2\times 2^5\times q^2$. So, the prime divisors of $f(6)$ belong to the set $\{3,p\}$ and als...
- Thu Jan 19, 2017 8:00 pm
- Forum: Secondary Level
- Topic: Dhaka Regional '16 P8
- Replies: 3
- Views: 3499
Re: Dhaka Regional '16 P8
No. Experimenting with smaller values yield $2^n-n-1$ as a conjecture and it can be finished off with a recursion argument.
- Thu Jan 19, 2017 3:53 pm
- Forum: Algebra
- Topic: Seems ugly, But cute
- Replies: 5
- Views: 3918
Re: Seems ugly, But cute
Thamim really outdone himself this time. In fact, a proof of cauchy-schwarz inequality relies on the expansion of the product, and in that sense while Thamim's solution might look lengthy, they are actually same in the core. Of course, I like the vector product proof of cauchy-schwarz inequality be...
- Thu Jan 19, 2017 1:15 pm
- Forum: Secondary Level
- Topic: Dhaka Regional '16 P8
- Replies: 3
- Views: 3499
Re: Dhaka Regional '16 P8
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