Search found 75 matches
- Wed Jan 15, 2014 10:22 am
- Forum: Divisional Math Olympiad
- Topic: Barisal Secondary 2013 / 8
- Replies: 12
- Views: 8654
Re: Barisal Secondary 2013 / 8
আপু এক্টু বুঝায় দেন। মাথায় ঢুকসে না।
- Mon Jan 13, 2014 6:36 pm
- Forum: Divisional Math Olympiad
- Topic: Some problems of 2013
- Replies: 1
- Views: 2147
Some problems of 2013
I need some help with some problems of 2013 Divisional BDMO. 2013 Barisal Secondary $$8.$$ The incenter of triangle $$ABC$$ is $$I$$ and inradius is $$2$$. What is the smallest possible value of $$AI+BI+CI$$ ? $$9.$$From the $16$ lattice points of the $3*3$ grid, you have to choose $6$ so that at le...
- Mon Jan 13, 2014 5:20 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 8, Higher Secondary 6
- Replies: 8
- Views: 12070
Re: BdMO National 2013: Secondary 8, Higher Secondary 6
@Fatin, Not Clear to me at all! Faulty Solution. First, try to explain it clearly. I was trying to tell that it is not possible to come back to its original position without ever traveling the same road twice. Let there be $$C$$ ways to choose $$n$$ roads such that we can not retun to its original ...
- Mon Jan 13, 2014 5:01 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 5
- Replies: 4
- Views: 5135
Re: BdMO National 2013: Secondary 5
There was a mistake Let P and Q be two points on EF and M and N be two points on BC such that EP=FQ=DM=CN=OG. Now $$(PMNQ)= 1/2 * (PQ+MN)*OG=1/2 * (100-25\sqrt{3} +75- 25\sqrt{3})*25\sqrt{3}/2 $$ $$=(175-50\sqrt{3})*25\sqrt{3}/4 $$ $$(ABCD)= 1/2 * (AB+CD)* OA = 1/2 * (100+50)* 25\sqrt{3}$$ $$= 150*2...
- Sun Jan 12, 2014 9:07 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 8, Higher Secondary 6
- Replies: 8
- Views: 12070
Re: BdMO National 2013: Secondary 8, Higher Secondary 6
Let therebe C ways to take n points such that it is not possible to return to its original position. So, we can take 1 point in $$(n-1)$$. 2 points be taken in $$(n-1)(n-2)$$ ........... So $$n$$ points can be taken $$(n-1)(n-2)(n-3) ........(n-n+1)(n-n)$$ $$C= (n-1)(n-2)........(n-n+1)(n-n) =0$$. S...
- Sun Jan 12, 2014 7:41 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 5
- Replies: 4
- Views: 5135
Re: BdMO National 2013: Secondary 5
Joining $$A,C$$ $$AC^2=AB^2+BC^2-2AB.BCcos120= 50^2+ 50^2+50*50=3*50^2$$ $$AC=50\sqrt{3}$$. $$\angle BAC= \angle ACB=30^\circ, \angle BAC=90^\circ, \angle ACB=30^\circ$$. $$CD=AC/sin60 = 100$$, $$AD=CD/sin30 = 50$$. Let E and F be the midpoint of AD and BC. $$EF= (AB+CD)/2= 75$$. Let OA be the perpe...
- Sun Jan 12, 2014 5:50 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 9, Higher Secondary 7
- Replies: 3
- Views: 4374
Re: BdMO National 2013: Secondary 9, Higher Secondary 7
If $$p=2$$ then $$p+2q \equiv 0 (mod 2)$$ So $$p \neq 2$$. If $$p=3$$ then $$2q=2,4$$ So, $$p+2q=5,7$$ So, $$3$$ is a awsome prime. If $$p>3$$ then there are q's such that $$2q \equiv 0,1,2 (mod 3)$$. Now, if $$p \equiv 1,2(mod 3)$$ So $$p+2q \equiv 0 (mod 3) $$ for some q's. So $$p \ngtr 3$$. So $...
- Sun Jan 12, 2014 5:32 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 2, Higher Secondary 1
- Replies: 4
- Views: 8365
Re: BdMO National 2013: Secondary 2, Higher Secondary 1
As $$P$$ and $$Q$$ are the midpoints of $$AB$$ and $$AE$$, $$AB=AE$$. $$\therefore AP=AQ, PQ||BE$$. But $$PQ||CD$$. $$\therefore BE||CD$$. $$\therefore \angle APQ=\angle AQP=\angle ABE=\angle AEB=\angle ACD$$. $$\angle ABE+ \angle EBD= 90^\circ= \angle BED + \angle EBD =>\angle ABE=\angle BED $$ In...
- Sun Jan 12, 2014 5:24 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 1
- Replies: 1
- Views: 3348
- Fri Jan 03, 2014 3:17 pm
- Forum: Secondary Level
- Topic: Prove it!
- Replies: 0
- Views: 1894
Prove it!
Given triangle $ABC$, points $P, Q, R$ are taken on $BC, CA, AB$ so that $BP:PC=CQ:QA=AR:RB=m:n$.
Prove that area $(PQR)$ : area $(ABC)=m^3+n^3:(m+n)^3$
Prove that area $(PQR)$ : area $(ABC)=m^3+n^3:(m+n)^3$