Search found 23 matches
- Fri Aug 30, 2013 9:05 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 6 (EXAM!)
- Replies: 36
- Views: 28199
Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)
I have sent you email with 6 attachments. Hope you received it.
- Thu Aug 29, 2013 10:42 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 5 (EXAM!)
- Replies: 29
- Views: 24179
Re: [OGC1] Online Geometry Camp: Day 5 (EXAM!)
নায়েল ভাইয়া, আমি প্রথমে PM পাঠাতে চেষ্টা করেছিলাম। কিন্তু কোন কারণে পারছিলাম না। তাই পরে মেইল করেছি। কিন্তু, যেভাবে লিখেছি, equation গুলো ওভাবে যায়নি। Sorry ভাইয়া, একটু কষ্ট করে পড়ে নিবেন please.
- Thu Aug 29, 2013 9:32 am
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 5 (EXAM!)
- Replies: 29
- Views: 24179
Re: [OGC1] Online Geometry Camp: Day 5 (EXAM!)
I have installed Latex fonts, but still i can't use it. What should i do?
- Wed Aug 28, 2013 10:52 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 4
- Replies: 32
- Views: 26941
Re: [OGC1] Online Geometry Camp: Day 4
well, yes!photon wrote:@ Samiun Fateeha Ira , apply power of a point any other way.
By the way , did it occur to you that $AB=48$ ?
AB= AD+DE+EB= 3+39+6= 48.
- Wed Aug 28, 2013 10:50 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 4
- Replies: 32
- Views: 26941
Re: [OGC1] Online Geometry Camp: Day 4
Mursalin wrote:You can click on the original-problem-link and see the solution but I don't recommend that. It's not good to give up on a problem that easily!
Everything you did is correct. And you shouldn't get multiple solutions. Would you like a hint?
i don't want to see the solution. just give a hint.
- Wed Aug 28, 2013 9:37 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 4
- Replies: 32
- Views: 26941
Re: [OGC1] Online Geometry Camp: Day 4
problem 01 :
BE.BD= BF.BG
or, 6*(39+6)=BF(BF+21)
or, BF^2+21BF-270=0
so, BF=9
so, CG=48-21-9=18
suppose, AI=x, HI=y, CH=z
Now, i have 3 equations:
x(x+y)=126, z(y+z)=702, x+y+z=48
but, i am stucked here. i am getting a lot of solution sets. what should i do & where is my mistake?
BE.BD= BF.BG
or, 6*(39+6)=BF(BF+21)
or, BF^2+21BF-270=0
so, BF=9
so, CG=48-21-9=18
suppose, AI=x, HI=y, CH=z
Now, i have 3 equations:
x(x+y)=126, z(y+z)=702, x+y+z=48
but, i am stucked here. i am getting a lot of solution sets. what should i do & where is my mistake?
- Tue Aug 27, 2013 6:17 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 2
- Replies: 22
- Views: 19345
Re: [OGC1] Online Geometry Camp: Day 2
ওই জে...কাগজ ভাঁজের কথা বলা আছে না, তাই করে পেয়েছি।Neblina wrote:@ Ira
How did you prove this-
PQ রেখাংশ এবং AC কর্ণ পরস্পর O বিন্দুতে সমকোণে সমদিখণ্ডিত হয়।?
- Tue Aug 27, 2013 6:54 am
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 2
- Replies: 22
- Views: 19345
Re: [OGC1] Online Geometry Camp: Day 2
Well, I was also unable to see Bangla texts from Google chrome. but now i can read it browsing Mozilla.
- Tue Aug 27, 2013 6:51 am
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 2
- Replies: 22
- Views: 19345
Re: [OGC1] Online Geometry Camp: Day 2
ধরি, $P$ বিন্দু $AD$ এর ওপর এবং $Q$ বিন্দু $BC$ এর ওপর অবস্থিত। $PQ$ রেখাংশ এবং $AC$ কর্ণ পরস্পরকে $O$ বিন্দুতে সমকোণে সমদ্বিখণ্ডিত করে। $AC^2= AD^2+ DC^2= 25+144=169\Longrightarrow AC= 13$ $AO=6.5$ $\triangle AOP$ ও $\triangle ADC$ এ, $\angle AOP=\angle ADC$, $\angle OAP=\angle DAC$. সুতরাং, $\tria...
- Mon Aug 26, 2013 9:12 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 2
- Replies: 22
- Views: 19345
Re: [OGC1] Online Geometry Camp: Day 2
$AD$ bisects $\angle BAC$. So, $\dfrac {AB}{BC}= \dfrac{BD}{DC}$ $\sqrt {\dfrac {41}{164}} = \dfrac {BD}{DC}\Longrightarrow \dfrac 1 2=\dfrac{BD}{DC}\Longrightarrow BD= \dfrac 1 2 DC$. From Stewart's theorem, in triangle ABC, \begin{align*} & AC^2\cdot BD+AB^2\cdot DC= BC(AD^2+ BD\cdot DC)\\ & \frac...