## Search found 172 matches

- Fri May 27, 2011 7:21 am
- Forum: Geometry
- Topic: cyclic quads to parallelogram
- Replies:
**5** - Views:
**2273**

### Re: cyclic quads to parallelogram

Please... someone try to to solve the problem (I am asking junior and secondary level students). There is a hint about the problem: Draw the figure using ruler and compass. Surely you will find something interesting. Not only for this problem, after drawing the figure ( properly ) of almost any geom...

- Wed May 25, 2011 9:23 am
- Forum: Higher Secondary Level
- Topic: finding the equation from graph..
- Replies:
**4** - Views:
**2339**

### Re: finding the equation from graph..

Keu ki bolte parba jodi kono somikoroner lekhor koekta bindu deoa thake ,tahole ar theke somikoron ta bair kora jabe ki...jemon(2,3),(5,8) ar somikoron ki hobe... :?: You have to be given enough points to be sure about the graph. From only two points, you can't be sure about the equation. For examp...

- Wed May 25, 2011 7:57 am
- Forum: Number Theory
- Topic: about harmonic means
- Replies:
**3** - Views:
**1609**

### Re: about harmonic means

The first thing I found interesting about HM is - 'If you travel same distance at $a_1$ speed, then at $a_2$ speed,...,then at $a_n$ speed, your mean speed will be the HM of $a_1,a_2,...,a_n$. Not the AM. Also if the distance doesn't remain same, you can use weighted HM to find the mean speed.'

- Wed May 11, 2011 8:21 pm
- Forum: Geometry
- Topic: Secondary Special Camp 2011: Geometry P 4
- Replies:
**6** - Views:
**11503**

### Re: Secondary Special Camp 2011: Geometry P 4

$M$ is the midpoint of side $BC$. $\angle AQD=\angle AEF=\angle B$ and $\angle ARD=\angle AFE=\angle C$ \[\angle EPB=180\circ - \angle PEB-\angle EBP=180\circ-(\angle AEB-\angle AEF)-(\angle EBA+\angle ABP)\] \[=180\circ-(90\circ-\angle B)-\{(90\circ-\angle A)+(180\circ-\angle B)\}=2\angle B+\angle ...

- Wed May 11, 2011 3:14 pm
- Forum: Geometry
- Topic: Secondary Special Camp 2011: Geometry P 3
- Replies:
**3** - Views:
**2309**

### Re: Secondary Special Camp 2011: Geometry P 3

Let Extended $AP$ meets $BC$ at $D'$. As, $AC=CD,\ \angle ADC=\angle CAD$ $\angle CPD'=180\circ -\angle APC=\angle ADC$ $\angle CPD=\angle CAD$ $\Rightarrow \angle CPD=\angle CPD'$ $\therefore PC$ is the internal bisector of $\angle D'PD$ As, $\angle BPC=90\circ,\ PB$ is the external bisector of $\a...

- Wed May 11, 2011 7:49 am
- Forum: Number Theory
- Topic: Secondary Special Camp 2011: NT P 3
- Replies:
**5** - Views:
**2530**

### Re: Secondary Special Camp 2011: NT P 3

Thank you

- Tue May 10, 2011 4:14 pm
- Forum: Number Theory
- Topic: Secondary Special Camp 2011: NT P 3
- Replies:
**5** - Views:
**2530**

### Re: Secondary Special Camp 2011: NT P 3

... $\text{Lemma}$ : It is a very useful one in number theory. Every odd prime divisor of $n^2+1$ is of the form $4k+1$. $\text{ Proof}$ : $n^2\equiv-1\pmod p\Longrightarrow n^4\equiv1\pmod p$. Also, by Fermat's little theorem, $n^{p-1}\equiv1\pmod p$. Thus, $4|p-1\Longrightarrow p\equiv1\pmod 4$. ...

- Tue Apr 19, 2011 11:58 pm
- Forum: Geometry
- Topic: DRAW A MAP BY FOUR COLOURS (am i solve it ?)
- Replies:
**3** - Views:
**2010**

### Re: DRAW A MAP BY FOUR COLOURS (am i solve it ?)

I also haven't understand some parts. Here is my first question: Probably you meant something like this in the second para of your proof: desh.JPG Then what should I do if first and third neighboring country have same border? You said alternatively we will colour the first point with C2, second with...

- Wed Mar 30, 2011 11:01 am
- Forum: Geometry
- Topic: Problem! Problem!
- Replies:
**8** - Views:
**3229**

### Re: Problem! Problem!

Let $\beta=\angle BAD\ (=\angle ABD)$ and $\gamma=\angle CAE\ (=\angle AEC)$ $A=\beta + \gamma$ $\angle DEF=\angle CAE + \angle AEC = 2 \gamma$ $\angle FDE=\angle BAD + \angle ABD = 2 \beta$ $\therefore \angle BFC = \angle DEF + \angle FDE = 2\gamma + 2 \beta = 2(\beta+\gamma) = 2A$ $\therefore$ ref...

- Tue Mar 29, 2011 9:04 am
- Forum: Geometry
- Topic: Game of pool? (own)
- Replies:
**6** - Views:
**2933**

### Re: Game of pool? (own)

@Mahi.. Thanks for the diagram. I did it in same way. I skipped the calculation part of your solution. You assumed that $2AB=BC$ but the problem stated $AB=2BC$. That's not a problem. But probably you have considered holes in the middle of shorter sides. But there are holes in the middle of longer s...