Search found 110 matches
- Sun Jan 12, 2014 2:14 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Junior 8
- Replies: 3
- Views: 3947
Re: BdMO National 2013: Junior 8
ok labib vaia ...... i'll try that .....
- Sun Jan 12, 2014 1:54 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Junior 4
- Replies: 1
- Views: 3125
Re: BdMO National 2013: Junior 4
2 divides a, but 4 doesn't . so a is form as $a=2(2k+1)$ so, $a^{2012}+a^{2013}+a^{2014}+.........+a^{3012}$ $=\left \{ 2(2k+1) \right \}^{2012}+\left \{ 2(2k+1) \right \}^{2013}+\left \{ 2(2k+1) \right \}^{2014}+.........+\left \{ 2(2k+1) \right \}^{3012} $ $=2^{2012}(2k+1)^{2012}+2^{2013}(2k+a)^{2...
- Sun Jan 12, 2014 12:38 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Junior 9
- Replies: 7
- Views: 13178
Re: BdMO National 2013: Junior 9
let two numbers are a and b. and let $\left ( a,b \right )=x \Leftrightarrow a=xa_{1}$ and $b=xb_{1}$ where $\left ( a_{1},b_{1} \right )=1$ so,$\left [ a,b \right ]=xa_{1}b_{1}$ but given that, \[ \frac{(a,b)}{[a,b]}=\frac{1}{36} \] \[\frac{x}{xa_{1}b_{1}}=\frac{1}{36}\Leftrightarrow \frac{1}{a_{1}...
- Sat Jan 11, 2014 11:57 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Junior 8
- Replies: 3
- Views: 3947
Re: BdMO National 2013: Junior 8
from pythagoras's theorem, $AC^{2}=AB^{2}+BC^{2}$ $AC^{2}=\sqrt{2}^{2}+\sqrt{6}^{2}$ $AC^{2}=2+6=8$ $AC=\sqrt{8}=2\sqrt{2}$ now for $\Delta ABC\Leftrightarrow sin\angle ACB=\frac{AB}{AC}=\frac{\sqrt{2}}{2\sqrt{2}}=\frac{1}{2}$ $\therefore \angle ACB=30$ now for $\Delta PRC\Leftrightarrow sin 30=\fra...
- Sat Jan 11, 2014 10:49 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Junior 3
- Replies: 1
- Views: 2935
Re: BdMO National 2013: Junior 3
$\Delta ANC\sim \Delta BMC$
$\frac{AN}{BM}=\frac{AC}{BC}$
$\frac{6}{BM}=\frac{12}{10}$
$BM=\frac{6*10}{12}=\frac{60}{12}=5$
$BM=5 \left ( ans: \right )$.......
$\frac{AN}{BM}=\frac{AC}{BC}$
$\frac{6}{BM}=\frac{12}{10}$
$BM=\frac{6*10}{12}=\frac{60}{12}=5$
$BM=5 \left ( ans: \right )$.......
- Thu Jan 09, 2014 7:35 pm
- Forum: Geometry
- Topic: This is not cool actually!
- Replies: 4
- Views: 3775
Re: This is not cool actually!
area of square is 2 ....... so, AC=2.
$\Delta ABC= 2^{2} \times \frac{\sqrt{3}}{4}= \sqrt{3}$ [becaus ABC is equilateral]
$\Delta ADC=\frac{2}{2}=1$
$ABCD= \sqrt{3}+1= 1+(1 \times \sqrt{3})= a+b\sqrt{c}$
$a+b+c= 1+1+3=5$ .......so ans:5
$\Delta ABC= 2^{2} \times \frac{\sqrt{3}}{4}= \sqrt{3}$ [becaus ABC is equilateral]
$\Delta ADC=\frac{2}{2}=1$
$ABCD= \sqrt{3}+1= 1+(1 \times \sqrt{3})= a+b\sqrt{c}$
$a+b+c= 1+1+3=5$ .......so ans:5
- Fri Aug 30, 2013 11:27 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 6 (EXAM!)
- Replies: 36
- Views: 28457
Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)
@nahraf solution of problem no:6 let the line through K parallel to BC meets the circumcircle at R. thus KP IS the reflection of KR respect to line KC.So,$\angle CKP=\angle CKR$ now,$\angle CKP=\angle CBP$ and $\angle CKR=\angle KCB$ $\therefore \angle CBP=\angle KCB$ or,$\angle CBP=90-\angle ABC$ o...
- Fri Aug 30, 2013 1:18 am
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 6 (EXAM!)
- Replies: 36
- Views: 28457
Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)
ভাইয়া,সমস্যা গুলোর সমাধান পাওয়ার পর reply দিয়ে জানিয়ে দিলে ভাল হয়।
- Fri Aug 30, 2013 12:53 am
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 5 (EXAM!)
- Replies: 29
- Views: 24409
Re: [OGC1] Online Geometry Camp: Day 5 (EXAM!)
vaia...for a problem in internet i tried but can't send my solutions timely.....so,don't mind and plz accept my solutions.
- Thu Aug 29, 2013 5:17 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 5 (EXAM!)
- Replies: 29
- Views: 24409
Re: [OGC1] Online Geometry Camp: Day 5 (EXAM!)
ভাইয়া 1 number সমস্যাটির চিত্রে ভাজ করার পর কাগজের কোনাটি বিপরীত বাহুর উপরে পড়বে......এটি বোঝানো হয়েছে?????......নাকি কোনাটি বাহুর বাইরে পড়বে বোঝানো হয়েছে?????