Search found 172 matches

by Zzzz
Fri May 27, 2011 7:21 am
Forum: Geometry
Topic: cyclic quads to parallelogram
Replies: 5
Views: 2194

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...
by Zzzz
Wed May 25, 2011 9:23 am
Forum: Higher Secondary Level
Topic: finding the equation from graph..
Replies: 4
Views: 2260

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...
by Zzzz
Wed May 25, 2011 7:57 am
Forum: Number Theory
Topic: about harmonic means
Replies: 3
Views: 1553

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.'
by Zzzz
Wed May 11, 2011 8:21 pm
Forum: Geometry
Topic: Secondary Special Camp 2011: Geometry P 4
Replies: 6
Views: 11388

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 ...
by Zzzz
Wed May 11, 2011 3:14 pm
Forum: Geometry
Topic: Secondary Special Camp 2011: Geometry P 3
Replies: 3
Views: 2242

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...
by Zzzz
Wed May 11, 2011 7:49 am
Forum: Number Theory
Topic: Secondary Special Camp 2011: NT P 3
Replies: 5
Views: 2462

Re: Secondary Special Camp 2011: NT P 3

Thank you :)
by Zzzz
Tue May 10, 2011 4:14 pm
Forum: Number Theory
Topic: Secondary Special Camp 2011: NT P 3
Replies: 5
Views: 2462

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, $4|p-1\Longrightarrow p\equiv1\pmod 4$. ...
by Zzzz
Tue Apr 19, 2011 11:58 pm
Forum: Geometry
Topic: DRAW A MAP BY FOUR COLOURS (am i solve it ?)
Replies: 3
Views: 1948

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...
by Zzzz
Wed Mar 30, 2011 11:01 am
Forum: Geometry
Topic: Problem! Problem!
Replies: 8
Views: 3129

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...
by Zzzz
Tue Mar 29, 2011 9:04 am
Forum: Geometry
Topic: Game of pool? (own)
Replies: 6
Views: 2838

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...