## Search found 185 matches

- Sat Mar 04, 2023 8:58 am
- Forum: National Math Camp
- Topic: AIME
- Replies:
**2** - Views:
**4584**

### Re: AIME

Every time Peter loses, he doesn't get to play in the next game. So we can (almost) make an one-to-one map between Peter's losing and not playing days. i,e if $x=$ number of games Peter lost, then $x = (22 + 20 + 32) - 22 - x \pm 1 \Rightarrow x=26$. So Peter played $26 + 22 = 48$ games.

- Sat Mar 04, 2023 8:01 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary 2020 P10
- Replies:
**5** - Views:
**5770**

### Re: BdMO National Secondary 2020 P10

First of all, $(20, 13)$ can only be reached from $(19, 13)$. Notice that we can only take turns at points where both $x$ and $y$ coordinates have prime values. Which means we only have to consider points with prime coordinates, leaving us with a $6\times 4$ grid (because there are $7$ and $5$ prime...

- Fri Jun 02, 2017 8:49 am
- Forum: Number Theory
- Topic: Determine n
- Replies:
**1** - Views:
**2442**

### Re: Determine n

Sketch:

- Mon May 15, 2017 11:24 am
- Forum: Algebra
- Topic: Nice and hard problem!
- Replies:
**1** - Views:
**5131**

### Re: Nice and hard problem!

From the first equation if $y=f(n)^2-458$, then $f(y)=2y^2-2.458^2$

We can rewrite it as $f(y)-2y^2+2.458^2=0$. Suppose $g(y)=f(y)-2y^2+2.458^2$. So $g(y)$ is a polynomial and it has infinite zeros (obviously $f(n)^2-458$ can take infinite values). So $g(y)$ must be zero for all $y$. The rest is easy.

We can rewrite it as $f(y)-2y^2+2.458^2=0$. Suppose $g(y)=f(y)-2y^2+2.458^2$. So $g(y)$ is a polynomial and it has infinite zeros (obviously $f(n)^2-458$ can take infinite values). So $g(y)$ must be zero for all $y$. The rest is easy.

- Mon May 15, 2017 11:06 am
- Forum: Algebra
- Topic: Inequality with abc = 1
- Replies:
**3** - Views:
**6250**

- Mon Apr 03, 2017 7:55 pm
- Forum: Social Lounge
- Topic: Math
- Replies:
**7** - Views:
**7810**

### Re: Math

Dunno about the first question but if you want to learn directly from Asif E Elahi, I suggest that you start following him on fb. See his timeline and like all of his posts, share them etc etc. Then he might notice you. Also, Asif e Elahi likes eager students. Knock him on messenger/tg and ask for ...

- Mon Apr 03, 2017 7:53 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #1
- Replies:
**52** - Views:
**39514**

### Re: BDMO Forum Mafia #1

বাচ্চারা অংক করতে যাও

- Mon Jan 09, 2017 2:23 pm
- Forum: Algebra
- Topic: 2015 regional Secondary no. 6 function algebra
- Replies:
**4** - Views:
**4236**

### Re: 2015 regional Secondary no. 6 function algebra

Suppose F(x)=ax+b , then F(F(x))=(a^2)x+ab+b if we compare 4x+3 with (a^2)x+ab+b $F(F(x))$ is not equal to $4x+3$. $F(F(x))=f(f(f(f(x))))$ from your definition of $F$. then we will get F(x)=2x+1 or F(x)=-2x-3 . Considering the next statement we can be sure that F(x)=2x+1 . How do you deduce that? A...

- Sun Jan 08, 2017 2:30 pm
- Forum: Algebra
- Topic: 2015 regional Secondary no. 6 function algebra
- Replies:
**4** - Views:
**4236**

- Sun Jan 08, 2017 4:10 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**107739**

### Re: Geometry Marathon : Season 3

Problem 10: Let $I$ be the incenter of $\triangle ABC$. The incircle touches $BC$ at $D$ and $K$ is the antipode of $D$ in $(I)$. Let $M$ be the midpoint of $AI$. Prove that $KM$ passes through the Feuerbach Point . We define some new points 1. $L$ is the midpoint of $BC$ 2. $N$ is the point where ...