## Search found 172 matches

Fri May 27, 2011 7:21 am
Forum: Geometry
Replies: 5
Views: 2256

### 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: 2329

### 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
Replies: 3
Views: 1600

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: 11477

### 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: 2294