Search found 98 matches
- Sun Feb 26, 2017 12:04 am
- Forum: Geometry
- Topic: IGO 2016 Elementary/4
- Replies: 8
- Views: 6583
Re: IGO 2016 Elementary/4
The fun thing(or cruel) about this problem is: You don't even have to do angle chasing in this problem. But when it is saying something about $90^o$, all we want is diameter. Through, use everything you learnt in class $6,7,8$. By that, you might solve the problem.
- Sat Feb 25, 2017 12:51 am
- Forum: Social Lounge
- Topic: সেঞ্চুরি ! সেঞ্চুরি !! সেঞ্চুরি !!!
- Replies: 10
- Views: 19699
Re: সেঞ্চুরি ! সেঞ্চুরি !! সেঞ্চুরি !!!
Congrats on $10th$ century.
- Fri Feb 17, 2017 1:39 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/4
- Replies: 11
- Views: 8644
Re: BdMO 2017 junior/4
By judging the 6,7,8 ..... This one is kinda hard.... Rather than 8,9.
- Fri Feb 17, 2017 1:34 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/10
- Replies: 2
- Views: 2821
BdMO 2017 junior/10
Whenever Avik gets a sequence, he multiplies every two distinct terms of that sequence, and then sums up these products to get the Hocus-pocus sum of the sequence. For example, the Hocus-pocus sum for the sequence $a, b, c, d$ is $ab + bc + ac + ad + bd + cd$. If Avik gets a sequence of $100$ terms,...
- Mon Feb 13, 2017 4:54 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO 2017 : Junior 8
- Replies: 8
- Views: 5146
Re: National BDMO 2017 : Junior 8
I am actually kicking me for not trying 8,9 in the exam. They both are easy.
- Sun Feb 12, 2017 11:46 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/4
- Replies: 11
- Views: 8644
Re: BdMO 2017 junior/4
And I don't know how i missed the part (b) in the actual exam. And i not posting this solution. Left for the others.
- Sun Feb 12, 2017 11:43 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/4
- Replies: 11
- Views: 8644
BdMO 2017 junior/4
(a) Can two consecutive numbers $n$ and $n-1$ both be divisible by $3$?
(b) Determine the smallest integer $n > 1$ such that $n^2(n-1)$ is divisible by $1971$. Note:
$1971 = 3^3 \times 73$.
(b) Determine the smallest integer $n > 1$ such that $n^2(n-1)$ is divisible by $1971$. Note:
$1971 = 3^3 \times 73$.
- Sun Feb 12, 2017 9:19 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO 2017 : Junior 8
- Replies: 8
- Views: 5146
Re: National BDMO 2017 : Junior 8
We know that O is the circumcenter. So, $OA=OB=OC$. And $DB=DO$. So, $ \angle DBO=\angle DOB=a$. Take D' such that $D'C= \frac {1}{3} BC$. So, $DO=DB=DD'=D'C$. $OB=OC$. So, $\angle OBC= \angle OCB=a$. So by $SAS$ congruence $\triangle OBD \cong \triangle OCD'$. So, $OD=OD'$, And, $\triangle ODD'$ is...
- Sun Feb 12, 2017 8:01 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/9
- Replies: 2
- Views: 2743
Re: BdMO 2017 junior/9
Here lets take $\angle DAP=\angle b$ So,$\angle PAB=90^o-b$. We know hat $\angle APB=90^o$ Or, $\angle ABP=\angle b$ So, $\angle PBC=90^o-b$ Now, $\triangle CBP$ is isosceles such that $CP=CB$. We draw the angle bisector $CM$ of $\angle BCP$. Here $CM \perp PB$. And $\angle BCM=\angle b$. Now by $AS...
- Sun Feb 12, 2017 7:49 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/9
- Replies: 2
- Views: 2743
BdMO 2017 junior/9
$ABCD$ is a square. Circle with diameter $AB$ and circle with center $C$ and radius $CB$ meet inside the square at $P$. Prove that $DP = \sqrt2 AP$.