## Search found 90 matches

Sun Mar 04, 2018 1:15 pm
Forum: Site Support
Topic: general Question
Replies: 14
Views: 16665

### Re: general Question

Sun Feb 18, 2018 10:01 am
Sun Feb 18, 2018 9:25 am
Sun Feb 18, 2018 8:28 am

I think Zawad sir has generously permitted all.
Nope, he hasn't.

Oh, sorry.
Fri Feb 16, 2018 11:44 am
Forum: Site Support
Topic: general Question
Replies: 14
Views: 16665

### Re: general Question

Fri Feb 16, 2018 1:53 am
Thank you, moderators. Should I invite my friends in this forum?
Sure feel free. The more the merrier.
Tue Feb 13, 2018 4:37 pm
Forum: Site Support
Topic: general Question
Replies: 14
Views: 16665

### Re: general Question

samiul_samin wrote:
Sun Dec 31, 2017 3:01 pm
My 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: 30168

### Re: jes 17 catb 10

Tue Jun 13, 2017 4:38 pm
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 68
Views: 19993

### 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: 9473

### Re: MPMS Problem Solving Marathon

aritra barua wrote:For problem 1,if we substitute $a^2$=$4k$ where $k$ is an integer,we can easily find that $p$ |$a^2$+$1$.
For this you need to have $4k$ a perfect square, which isn't true for all $k$.
Fri May 12, 2017 1:44 am
Forum: Algebra
Topic: FE: Brahmagupta-Fibonacci identity!!!
Replies: 1
Views: 3993

### 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: 111
Views: 56808

### 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.
Mon May 01, 2017 2:35 am
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 111
Views: 56808

### 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: 27119

### Re: BDMO Forum Mafia #2

Nobody died.

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