Search found 136 matches
- Mon Mar 02, 2015 9:10 pm
- Forum: Secondary Level
- Topic: Bdmo 2013 secondary
- Replies: 3
- Views: 6274
Re: Bdmo 2013 secondary
A graph with $n$ vertices and $n$ edges contains a cycle. (With $n-1$ edges and no cycles it must be a tree, so the $n$th edge forms a cycle.)
- Mon Mar 02, 2015 3:47 am
- Forum: Number Theory
- Topic: Some GCD Problems
- Replies: 6
- Views: 6470
Re: Some GCD Problems
.$a^{2^{n}}+1$ is a divisor of $a^{2^{m}}-1$. how ? :?: :?: $a=3$ ; $n=2$ ; $m=3$ breaks it . I can't see $(a,n,m)=(3,2,3)$ yielding a contradiction. In fact, if $A_n=a^{2^n}+1$ and $A_m=a^{2^m}-1$ and $m=n+j$ then \[a^{2^n}\equiv -1~\left(\bmod ~A_n\right)\Longrightarrow \left(a^{2^n}\right)^{2^j}...
- Sat Feb 28, 2015 8:00 pm
- Forum: Combinatorics
- Topic: Partial Generalization of ISL '94 C5
- Replies: 5
- Views: 4696
- Fri Feb 27, 2015 12:27 pm
- Forum: Geometry
- Topic: Bulgaria 1996
- Replies: 4
- Views: 3922
Re: Bulgaria 1996
WAT?Fm Jakaria wrote: ... We are finished.
- Tue Feb 24, 2015 1:10 pm
- Forum: Geometry
- Topic: Balkan MO 2005
- Replies: 3
- Views: 3126
Re: Balkan MO 2005
All the angles are directed $\bmod ~\pi$ since there can be $2/3$ distinct configurations. $I$ is the incenter. Since $AD=AE$ we have $\measuredangle DEA = \measuredangle ADE$. Then notice that \[\begin{eqnarray} \measuredangle DXI &=&\measuredangle DXC=\measuredangle EXC=\measuredangle ECX+\measure...
- Mon Feb 23, 2015 11:16 pm
- Forum: Secondary Level
- Topic: (3n+1)^2+4n^3=m^2
- Replies: 3
- Views: 3557
Re: (3n+1)^2+4n^3=m^2
For all $k\in\mathbb N$ we have $n=k^2+k\Rightarrow (3n+1)^2+4n^3=(2k+1)^2(k^2+k+1)^2$.
- Mon Feb 23, 2015 3:54 pm
- Forum: Geometry
- Topic: Italy TST 2000/2
- Replies: 3
- Views: 3152
Re: Italy TST 2000/2
The key move is to prove that $AE\perp DM$. Then $DCEF$ is cyclic with $CD = CE$ and the result follows. Let $A\equiv (0,6),B\equiv (6,0),C\equiv (0,0)$ , then $M\equiv(3,3),D\equiv(0,2),E\equiv(2,0)$. We quickly get $\overleftrightarrow{AE}\equiv 3x+y=6$ and $\overleftrightarrow{DM}\equiv 3y-x=6$. ...
- Sun Feb 22, 2015 1:59 am
- Forum: Higher Secondary Level
- Topic: Vectors around Regular Polygon
- Replies: 4
- Views: 13064
Re: Vectors around Regular Polygon
We choose the coordinate system so that $\overrightarrow{OP_1}$ is parallel to $x$-axis and pointed to the positive direction. Breaking each vector into components we get $\overrightarrow{OP_i}=r\cos\dfrac{2\pi(i-1)}{n}\hat{\mathbf{i}}+r\sin\dfrac{2\pi(i-1)}{n}\hat{\mathbf{j}}$ so \[\sum_{i=1}^n \ov...
- Wed Feb 18, 2015 4:44 pm
- Forum: Combinatorics
- Topic: Iran 3rd round 2013
- Replies: 4
- Views: 3852
Re: Iran 3rd round 2013
How to apply Jensen? Please explain. I think Cauchy Schwarz does the work. Suppose there are $a_k$ rooks on $k$th row and $b_k$ rooks on $k$th column. Then a rook on row $i$ is threatened by $\left(a_i-1\right)$ rooks (all excluding itself). Similarly a rook on column $i$ is threatened by $\left(b_...
- Mon Feb 09, 2015 1:08 pm
- Forum: Algebra
- Topic: factorial vs powers
- Replies: 1
- Views: 2239
Re: factorial vs powers
If $n,n(r)\in\mathbb N$, I think it suffices to show that $\sqrt[n]{n!}$ is increasing, so eventually it will get past any constant $r$. Indeed we have $\sqrt[n]{n!}<\sqrt[n+1]{(n+1)!}\Leftrightarrow n!^{n+1}<(n+1)!^n\Leftrightarrow n!<(n+1)^n$ which is obviously true for all $n$.