Yes, Zawad has. I dunno who Zawad sir is.M. M. Fahad Joy wrote: ↑Sun Feb 18, 2018 10:01 am

Oh, sorry.

## Search found 90 matches

- Sun Mar 04, 2018 1:15 pm
- Forum: Site Support
- Topic: general Question
- Replies:
**14** - Views:
**19773**

### Re: general Question

- Fri Feb 16, 2018 11:44 am
- Forum: Site Support
- Topic: general Question
- Replies:
**14** - Views:
**19773**

### Re: general Question

Sure feel free. The more the merrier.M. M. Fahad Joy wrote: ↑Fri Feb 16, 2018 1:53 amThank you, moderators. Should I invite my friends in this forum?

- Tue Feb 13, 2018 4:37 pm
- Forum: Site Support
- Topic: general Question
- Replies:
**14** - Views:
**19773**

### Re: general Question

Your posts should be approved by now. Sorry about the delay.samiul_samin wrote: ↑Sun Dec 31, 2017 3:01 pmMy posts are not published!!!It said the moderator need to watch my post first and then it will bw published.How can I see my post?

- Tue Feb 13, 2018 4:34 pm
- Forum: Geometry
- Topic: jes 17 catb 10
- Replies:
**2** - Views:
**31275**

### Re: jes 17 catb 10

Could you please elaborate?

- Tue Jun 13, 2017 4:38 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**26746**

### Re: Beginner's Marathon

An easy problem:$\text{Problem 20}$ The fractions $\frac{7x+1}{2}$, $\frac{7x+2}{3}$,......., $\frac{7x+2016}{2017}$ are irreducible.Find all possible values of $x$ such that $x$ is less than or equal to $300$.This Problem was recommended by Zawad bhai in a mock test at CMC. Here's a hint for you g...

- Mon May 29, 2017 5:02 pm
- Forum: News / Announcements
- Topic: MPMS Problem Solving Marathon
- Replies:
**11** - Views:
**12573**

### Re: MPMS Problem Solving Marathon

For this you need to have $4k$ a perfect square, which isn't true for all $k$.aritra barua wrote:For problem 1,if we substitute $a^2$=$4k$ where $k$ is an integer,we can easily find that $p$ |$a^2$+$1$.

- Fri May 12, 2017 1:44 am
- Forum: Algebra
- Topic: FE: Brahmagupta-Fibonacci identity!!!
- Replies:
**1** - Views:
**4430**

### Re: FE: Brahmagupta-Fibonacci identity!!!

No one posted any solution. Sillies. Let $P(x,y,z,t)$ denote the statement that $[f(x)+f(y)][f(z)+f(t)]=f(xz+yt)+f(xt-yz) $ Now, $P(0,0,0,0) \Rightarrow 2f(0)(2f(0) - 1) = 0$ So we have $f(0)=\dfrac{1}{2}$ or $f(0)=0$ Suppose $f(0)=\dfrac{1}{2}$ . Then, $P(x,0,0,0) \Rightarrow f(x)=\dfrac{1}{2}$ We ...

- Mon May 01, 2017 3:10 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**75310**

### Re: Geometry Marathon : Season 3

$\text{Problem 41}$

Let $ABC$ be a triangle, and $I$ the incenter, $M$ midpoint of $ BC $, $ D $ the touch point of incircle and $ BC $. Prove that perpendiculars from $M, D, A $ to $AI, IM, BC $ respectively are concurrent.

Let $ABC$ be a triangle, and $I$ the incenter, $M$ midpoint of $ BC $, $ D $ the touch point of incircle and $ BC $. Prove that perpendiculars from $M, D, A $ to $AI, IM, BC $ respectively are concurrent.

- Mon May 01, 2017 2:35 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**75310**

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 40:}$ Since $N$ is the Nagel point, we can easily prove that $BE=CD$ i.e. $BM=CT$. $P$ is the center of spiral similarity which sends $BE$ to $CD$ thus it also sends $M$ to $T$. Since $BE=CD$, we have the dilation factor at $1$ thus $PM=PT$. Now since a spiral similarity c...

- Wed Apr 26, 2017 8:22 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #2
- Replies:
**30** - Views:
**32067**

### Re: BDMO Forum Mafia #2

Nobody died.

Day 2 begins now and will end on 28th April, 8 pm.

Day 2 begins now and will end on 28th April, 8 pm.