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...

- Tue Jun 13, 2017 4:38 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**2802**

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 $k$.

- Mon May 29, 2017 5:02 pm
- Forum: News / Announcements
- Topic: MPMS Problem Solving Marathon
- Replies:
**8** - Views:
**428**

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 ...

- Fri May 12, 2017 1:44 am
- Forum: Algebra
- Topic: FE: Brahmagupta-Fibonacci identity!!!
- Replies:
**1** - Views:
**172**

$\text{Problem 41}$

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 3:10 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**90** - Views:
**3232**

$\text{Solution to Problem 40:}$ 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...

- Mon May 01, 2017 2:35 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**90** - Views:
**3232**

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.

- Wed Apr 26, 2017 8:22 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #2
- Replies:
**27** - Views:
**1113**

Nobody died, yay!

Night 1 starts now, will end at 26th April 8 pm.

I hope the mafia remember who they are, cause I totally forgot

Night 1 starts now, will end at 26th April 8 pm.

I hope the mafia remember who they are, cause I totally forgot

- Tue Apr 25, 2017 8:05 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #2
- Replies:
**27** - Views:
**1113**

Since this is dying, gonna fix a final deadline.

Day 1 will end on 25th, 8 pm. Discuss, do stuff, act!

Day 1 will end on 25th, 8 pm. Discuss, do stuff, act!

- Mon Apr 24, 2017 8:06 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #2
- Replies:
**27** - Views:
**1113**

We use barycentric coordinates. Let $P\equiv (p:q:r)$. Now, we know that $pq+qr+rp=0$ [The equation of circumcircle for equilateral triangles]. Now, $D\equiv (0:q:r), E\equiv (p:0:r), F\equiv (p:q:0)$. So, the area of $\triangle DEF$ divided by the area of $\triangle ABC$ is: $$\dfrac{1}{(p+q)(q+r)...

- Sat Apr 22, 2017 6:48 pm
- Forum: Geometry
- Topic: USA(J)MO 2017 #3
- Replies:
**6** - Views:
**293**

Bump bump bumpity bump.

How much time should I extend for?

How much time should I extend for?

- Sat Apr 22, 2017 8:27 am
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #2
- Replies:
**27** - Views:
**1113**