Search found 66 matches
- Fri Feb 03, 2017 11:57 am
- Forum: Geometry
- Topic: BDMO regional 2015
- Replies: 4
- Views: 4567
Re: BDMO regional 2015
$\text{Problem 8}$ Join $B,O$. In $\bigtriangleup ABO$ and $\bigtriangleup BOC$, $\angle COB = \angle ABO$ and $\angle CBO = \angle AOB$ and $BO$ is the common side. So, $\bigtriangleup ABO$ is congruent to $\bigtriangleup BOC$. So, $\angle OAB = \angle OCB \Rightarrow \angle OAB = \angle CBO \Right...
- Thu Feb 02, 2017 1:01 pm
- Forum: Secondary Level
- Topic: Find the angle
- Replies: 2
- Views: 3057
Find the angle
In triangle $\bigtriangleup ABC$, $\angle BCA = 70, \angle BAC = 30$. $M$ is a point inside the triangle such that $\angle ACM = 30, \angle CAM = 10$. Find $\angle BMC$.
- Thu Feb 02, 2017 11:52 am
- Forum: Junior Level
- Topic: BDMO National Junior 2016/6
- Replies: 9
- Views: 10711
Re: BDMO National Junior 2016/6
$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$ Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$ Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$ $\Rightarrow 3^{2w} ( 1 + 3^{3x-2w} + 3^{5y-2w}) = 3^{7z}$ $\Rightarrow 1 + 3^{3x-2w} + 3^{5y-2w} = 3^{7z-2w}$ Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3...
- Thu Feb 02, 2017 12:09 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies: 17
- Views: 11206
Re: BdMO National Secondary: Problem Collection(2016)
$\text{PROBLEM 5 (a)}$
- Tue Jan 31, 2017 11:07 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO national 2016, junior 10
- Replies: 7
- Views: 5546
Re: BDMO national 2016, junior 10
Cheekyahmedittihad wrote:I knew this problem was pretty much bland and disgusting. So, I just copied the solution from AOPS. Just copying took me 15 minutes. Allah knows how much time it took for the person to write.
- Tue Jan 31, 2017 10:19 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO national 2016, junior 10
- Replies: 7
- Views: 5546
Re: BDMO national 2016, junior 10
Note that this is equivalent to $a,b,c,d\mid a+b+c+d$ or $\text{lcm}(a,b,c,d)\mid a+b+c+d$. Letting $\text{lcm}(a,b,c,d)=X$, we want, $\dfrac{a}{X}+\dfrac{b}{X}+\dfrac{c}{X}+\dfrac{d}{X}=\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=k\in\mathbb{Z}^+$. The denominators are all integers...
- Tue Jan 31, 2017 12:35 am
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL Junior 2016/04
- Replies: 8
- Views: 5964
Re: BDMO NATIONAL Junior 2016/04
Let's assume $AC$ is not perpendicular to $BD$. WLOG, lets assume $\angle AOB < 90$. Let $X$ be the foot of the perpendicular from $B$ to AC and let $Y$ be the foot of the perpendicular from $D$ to $AC$. Now, extension of Pythagporous theorem gives $BO^2 + AO^2 + 2.AO.OX + DO^2 + CO^2+ 2.CO.OY = BO^...
- Sat Jan 28, 2017 11:46 pm
- Forum: Divisional Math Olympiad
- Topic: Divisional MO 2015
- Replies: 4
- Views: 3633
Re: Divisional MO 2015
See, $9800 = 2^3 \times 5^2 \times 7^2$ Let the two numbers will be $p = 2^a \times 5^b \times 7^c$ and $q = 2^x \times 5^y \times 7^z$. Now consider the pairs of $(a,x)$ which are possible: $(0,3), (1,3),(2,3),(3,3), (3,0), (3,1), (3,2),(1,2),(2,1)$ , numbers of pairs is $9$. Again, similarly the ...
- Sat Jan 28, 2017 4:17 pm
- Forum: Junior Level
- Topic: BDMO NATIONAL Junior 2016/03
- Replies: 4
- Views: 4188
Re: BDMO NATIONAL Junior 2016/03
If we mark our total diagonals as M,we find M=C(2016,2)-2016.Then our required diagonals will be less than the total number of diagonals.By cutting out a pattern.....we follow that for an N sided polygon,the required diagonals as per the question is 2n+(n-1)+(n-2).........+2+1...So,the required num...
- Thu Jan 26, 2017 11:18 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL 2016/03
- Replies: 2
- Views: 2250
Re: BDMO NATIONAL 2016/03
Answer