I can't see your solution.Ayantika rinti bose wrote: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 solution of problem 1 day2.‘half ’ because that is what i could do.
By angle bisector theorem we find that BD/DC=AB/AC.so,AB/DC =root 41/root 164=1/2.so,2BD=DC.
that’s that i could do.BUT PLZ NOTIFY ME IF YOU ARE SEEING MY SOLUTIONS!!!!!
[OGC1] Online Geometry Camp: Day 2

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Re: [OGC1] Online Geometry Camp: Day 2
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Re: [OGC1] Online Geometry Camp: Day 2
I think they are in different topics. Here are the total 4 posts you made  http://www.matholympiad.org.bd/forum/se ... 6&sr=posts . Did you make any other post than these?Tusher Chakraborty wrote:I can't see your solution.Ayantika rinti bose wrote: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 solution of problem 1 day2.‘half ’ because that is what i could do.
By angle bisector theorem we find that BD/DC=AB/AC.so,AB/DC =root 41/root 164=1/2.so,2BD=DC.
that’s that i could do.BUT PLZ NOTIFY ME IF YOU ARE SEEING MY SOLUTIONS!!!!!
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Nur Muhammad Shafiullah  Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi

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Re: [OGC1] Online Geometry Camp: Day 2
হি হি হি। আমারও চোরাই পদ্ধতিতে( প্র্যাক্টিক্যাল করে ) ৫.৪ উত্তর আসছে। থিওরিটিক্যালি পারতেসি না।