how fool i am . i did it by ceva with a lot of manipulation .

but tanmoy, your solution is quite easy .

## Search found 110 matches

- Thu Jul 09, 2015 12:19 am
- Forum: Geometry
- Topic: Torricelli's point
- Replies:
**2** - Views:
**1171**

- Wed Jul 08, 2015 1:21 pm
- Forum: Geometry
- Topic: Torricelli's point
- Replies:
**2** - Views:
**1171**

### 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:
**3684**

### 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$)

$=\frac{(m-1)m(2m-1)}{6}$

$={1}^{2}+{2}^{2}+......+{(m-1)}^{2}$

$\frac{n}{3}{m \choose 2}$

$=\frac{(2m-1)}{3}\frac{m(m-1)}{2}$ (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:
**3684**

### 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:
**8011**

### Re: Sylhet - 2014

not $3k$ . it was ${3}^{k}$ in the main problem .Mahfuz Sobhan wrote: Here x is divisible by 3k, if

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:
**2935**

### Re: USSR MO

this is obvious by euler's theorem .

so , skip this and try in another way .

so , skip this and try in another way .

- Tue Jul 07, 2015 1:58 am
- Forum: Secondary Level
- Topic: USSR MO
- Replies:
**3** - Views:
**2935**

### 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 .

{$ {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:
**3249**

### 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 $PB$ .

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:
**3249**

### Re: Iran NMO 2005/2

are you sure about the statement?

this line is confusing.tanmoy wrote: $CX \parallel PM$.

- Wed Jun 24, 2015 2:55 am
- Forum: Geometry
- Topic: cyclic quad
- Replies:
**3** - Views:
**2179**

### Re: cyclic quad

it should be $\angle C_{2}B_{1}Q$ , not $\angle C_{2}B_{1}P$ ..... typo ?tanmoy wrote:Now,$\angle C_{2}B_{1}P=\angle C_{2}AB=\angle C_{2}PB$