Vaia,

I have sent my solutions to nayel71@gmail.com.please check them.

Regards,

Ayantika.

## Search found 5 matches

- Fri Aug 30, 2013 8:42 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 6 (EXAM!)
- Replies:
**36** - Views:
**6215**

- Wed Aug 28, 2013 9:31 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 4
- Replies:
**32** - Views:
**5818**

### Re: [OGC1] Online Geometry Camp: Day 4

the area of large circle is 4 pi.and let the intersectors of the 4 small circles be P.then the area of 4 small circles area 4pi-4p.and let the upper shadow be R.then it is 4 pi-(4 pi-4p)=4r.so, it is p=r.so,a=0

- Wed Aug 28, 2013 8:22 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 2
- Replies:
**22** - Views:
**4157**

### Re: [OGC1] Online Geometry Camp: Day 2

Hello, Actually i am not so sure if you are seeing my solutions or not because you are not replying to me.i was ill so i joined the camp 2 days later,but i have done all the maths from day 1 and also replied 2 you.please atleast send me an email that you are seeing my solutions.and here is the half ...

- Wed Aug 28, 2013 11:43 am
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 1
- Replies:
**19** - Views:
**4745**

### Re: [OGC1] Online Geometry Camp: Day 1

Solution of ques. no 2 of day 1: In triangle $PQR$,$\angle PAQ=42$ and $AP=AQ$. Then,$\angle APQ=\angle AQP=69$.then $\angle EQT=\angle BPR=69$. So,in triangle $BQD$, $\angle BQD=111$. So $\angle QBD+\angle QDB=X+Y=69$. In triangle $BPR$ $\angle BPR=69$. So, $\angle PBR+\angle BRP=Y+Z=111$ Now we ha...

- Wed Aug 28, 2013 12:53 am
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 1
- Replies:
**19** - Views:
**4745**

### Re: [OGC1] Online Geometry Camp: Day 1

Hello bhaia, This is Ayantika Rinti Bose. Actually I was in the Lilaboyi camp till 24 august. And after coming home, I got severe cold and fever. That's why I couldn't join the online camp on 25th. But I wanted to do this barely. So now I am joining and giving my solution of question no 1. of day 1....