## Search found 110 matches

Thu Jul 09, 2015 12:19 am
Forum: Geometry
Topic: Torricelli's point
Replies: 2
Views: 1494

### Re: Torricelli's point

how fool i am . i did it by ceva with a lot of manipulation .
but tanmoy, your solution is quite easy .
Wed Jul 08, 2015 1:21 pm
Forum: Geometry
Topic: Torricelli's point
Replies: 2
Views: 1494

### Torricelli's point

Given a triangle $\Delta ABC$ . Let $\Delta ABF,\Delta BCD,\Delta CAE$ be equilateral triangles constructed outwards . Prove that $AD,BE,CF$ are concurrent .
Wed Jul 08, 2015 4:20 am
Forum: Secondary Level
Topic: Regular Polygon inscribed in a circle
Replies: 5
Views: 4341

### Re: Regular Polygon inscribed in a circle

i just linked up my solution with seemanta's one ,
$\frac{n}{3}{m \choose 2}$
$=\frac{(2m-1)}{3}\frac{m(m-1)}{2}$ (as$n=2m-1$)
$=\frac{(m-1)m(2m-1)}{6}$
$={1}^{2}+{2}^{2}+......+{(m-1)}^{2}$ seemanta001 wrote: $N^2+(N-1)^2+(N-2)^2+........+1$.
Wed Jul 08, 2015 4:09 am
Forum: Secondary Level
Topic: Regular Polygon inscribed in a circle
Replies: 5
Views: 4341

### Re: Regular Polygon inscribed in a circle

general solution: first choose any of $n$ points . now we have even numbers of rest points . draw a line with two vertices such that the line divides the rest points into equal parts. [the line must be parallel to one of the adjacent side of the choosen point] so one part has $\frac{n+1}{2}+1$ point...
Wed Jul 08, 2015 2:23 am
Forum: Secondary Level
Topic: Sylhet - 2014
Replies: 3
Views: 8676

### Re: Sylhet - 2014

Mahfuz Sobhan wrote: Here x is divisible by 3k, if
not $3k$ . it was ${3}^{k}$ in the main problem .
mahfuz sobhan you can use latex to avoid this type of mistakes.
Tue Jul 07, 2015 2:00 am
Forum: Secondary Level
Topic: USSR MO
Replies: 3
Views: 3421

### Re: USSR MO

this is obvious by euler's theorem . so , skip this and try in another way .
Tue Jul 07, 2015 1:58 am
Forum: Secondary Level
Topic: USSR MO
Replies: 3
Views: 3421

### USSR MO

Let $n\geq 3$ be an odd number . Show that there is a number in the set ,
{${2}^{1}-1,{2}^{2}-1,{2}^{3}-1,......,{2}^{n-1}-1$}
which is divisible by n .
Mon Jun 29, 2015 6:52 pm
Forum: Geometry
Topic: Iran NMO 2005/2
Replies: 3
Views: 3788

### Re: Iran NMO 2005/2

opsssss sorry my mistake

here is the solution:
$PM=YM$ [because $PM||XC$]
so,$KC=YM$

$\angle KBC=\angle BCY=\angle KCB$
$\therefore KB=KC$
$KB=KC=MY$

$\therefore MYBK$ is a parallelgram.
$\therefore \angle MKB=\angle MYB =90$
so, $MK$ is perpendicular to $PB$ .
Mon Jun 29, 2015 6:13 pm
Forum: Geometry
Topic: Iran NMO 2005/2
Replies: 3
Views: 3788

### Re: Iran NMO 2005/2

are you sure about the statement?
tanmoy wrote: $CX \parallel PM$.
this line is confusing. Wed Jun 24, 2015 2:55 am
Forum: Geometry
tanmoy wrote:Now,$\angle C_{2}B_{1}P=\angle C_{2}AB=\angle C_{2}PB$
it should be $\angle C_{2}B_{1}Q$ , not $\angle C_{2}B_{1}P$ ..... typo ?