## Search found 86 matches

- Wed Feb 01, 2017 2:24 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**16321**

### Re: Geometry Marathon : Season 3

Problem 18: Let $ABC$ be a triangle with circumcircle $\omega$ and let $H, M$ be orthocenter and midpoint of $AB$ respectively. Let $P,Q$ be points on the arc $AB$ of $\omega$ not containing $C$ such that $\angle ACP=\angle BCQ < \angle ACQ$.Let $R,S$ be the foot of altitudes from $H$ to $CQ,CP$ re...

- Wed Feb 01, 2017 1:02 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies:
**17** - Views:
**3348**

- Wed Feb 01, 2017 12:55 am
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO national 2016, junior 10
- Replies:
**7** - Views:
**1771**

### Re: BDMO national 2016, junior 10

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. :P Haha! :lol: It may be a well known NT problem. So sad that in bdmo national 2016 Junior Category's some ...

- Wed Feb 01, 2017 12:45 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies:
**17** - Views:
**3348**

### Re: BdMO National Secondary: Problem Collection(2016)

Problem 6(b): (b) $DO \cap BI $ $= E$. In $\triangle EIO,$ $\angle IOE + \angle EOI + \angle EIO = 180^{\circ}$. and also, $\angle EIO = \angle ICB+\angle CBI$ $\angle IEO = 180^{\circ} - ( \angle EOI + \angle EIO)$ $\dots (1)$ $AC = BC$ and $BIOD$ is a cyclic quadrilateral. $\angle EOI + \angle EI...

- Wed Feb 01, 2017 12:12 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies:
**17** - Views:
**3348**

### Re: BdMO National Secondary: Problem Collection(2016)

Problem 6(a): (a) NO, the lines $AC$ and $DI$ doesn't intersect because they are parallel. Claim 1: $I, O, C$ are collinear. Proof: $AC = BC$, So, $\angle BAC = \angle ABC$. O is the circumcenter of $\triangle$, $\angle OAB = \angle OBA$. So, $\angle OAC = \angle OBC$. Again, $\triangle OAC $ $\con...

- Tue Jan 31, 2017 11:42 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies:
**17** - Views:
**3348**

- Tue Jan 31, 2017 11:38 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies:
**17** - Views:
**3348**

### Re: BdMO National Secondary: Problem Collection(2016)

Problem 2 (a): (a) Factoring, $600 = 2^4 \times 3 \times 5^3$ $\therefore$ $600$ has $(4+1) (1+1) (3+1) = 40$ positive integers factor. Problem 2 (b): (b) Positive integer factors of $6000$ are perfect squares are $6$ in numbers. $\therefore$ $600$ has $(40 - 6) = 34$ positive integer factors which...

- Tue Jan 31, 2017 11:23 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies:
**17** - Views:
**3348**

### Re: BdMO National Secondary: Problem Collection(2016)

Problem 1(a): (a) Here, $n$, $n+1$, $n+2$ are there consecutive numbers. At least one of it will be even and at least one of it will be divisible by 3. So, in results, $6$ | $n(n+1)(n+2)$. Easy Solution: $\binom{n+2}{3}$ = $\frac{n(n+1)(n+2)}{3!}$ = $\frac{n(n+1)(n+2)}{6}.$ Problem 1(b): (b) $1^{20...

- Tue Jan 31, 2017 11:05 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies:
**17** - Views:
**3348**

### BdMO National Secondary: Problem Collection(2016)

Bangladesh National Mathematical Olympiad 2016 : Secondary Problem 1: (a) Show that $n(n + 1)(n + 2)$ is divisible by $6$. (b) Show that $1^{2015} + 2^{2015} + 3^{2015} + 4^{2015} + 5^{2015} + 6^{2015}$ is divisible by $7$. Problem 2: (a) How many positive integer factors does $6000$ have? (b) How ...

- Tue Jan 31, 2017 10:01 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO national 2016, junior 10
- Replies:
**7** - Views:
**1771**

### 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...