## Search found 181 matches

- Fri Sep 01, 2017 2:00 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**115** - Views:
**58397**

### Re: Geometry Marathon : Season 3

Define $K=(SPN)\cap (SQM)$ and let $X,Y$ denote the midpoints of $MQ,NP$ respectively. We will show that $XY||IO$, which proves the problem since it's well known in configurations pertaining to $MN=PQ$ that $XY||\ell$. By spiral similarity and $MN=PQ$, we have $KNM\cong KPQ$. Thus $KM=KQ\Rightarrow ...

- Wed Jul 19, 2017 12:25 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO $2017$ P$1$
- Replies:
**4** - Views:
**6734**

### IMO $2017$ P$1$

For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as $$a_{n+1} = \begin{cases} \sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\ a_n + 3 & \text{otherwise.} \end{cases} $$ Determine all values of $a_0$ so that there exists a number $A$ such that $a_n =...

- Tue Jul 04, 2017 11:53 pm
- Forum: Number Theory
- Topic: Infinite solutions
- Replies:
**1** - Views:
**1255**

### Re: Infinite solutions

$-a \equiv b^2 (moda+b^2) $.

So, $-a^3 \equiv b^6 (moda+b^2) $.

Which implies that $a+b^2 | b^6-b^3$.

Now fix $b$ and notice that for every divisor $x $ of $b^6-b^3$ there exists a $a $ such that $a+b^2=x $. So we get infinite solutions.

So, $-a^3 \equiv b^6 (moda+b^2) $.

Which implies that $a+b^2 | b^6-b^3$.

Now fix $b$ and notice that for every divisor $x $ of $b^6-b^3$ there exists a $a $ such that $a+b^2=x $. So we get infinite solutions.

- Fri Jun 30, 2017 12:31 pm
- Forum: Site Support
- Topic: help please!!!!!!!!
- Replies:
**1** - Views:
**7363**

### Re: help please!!!!!!!!

You upload attachments by clicking the 'Upload Attachment' and choosing a file to upload.

- Mon Jun 26, 2017 4:04 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**20947**

### Re: Beginner's Marathon

As this is the Beginner's Marathon, I request everyone to not give shortlist problems. Problem $25$ Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $ \Omega $ is a circle passing through $A,B,C,D$. Let $ \omega $ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ ...

- Mon Jun 26, 2017 3:58 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**20947**

### Re: Beginner's Marathon

Solution to Problem $24$ ( NOT OVERKILL ) As the previous solution we solve by contradiction. So, assume that $a_1a_k \equiv a_k (mod n)$. $ a_1\equiv a_1\cdot a_2\equiv a_1\cdot a_2\cdot a_3\equiv \dots \equiv a_1\cdot\dots\cdot a_k\equiv a_1\cdot\dots\cdot a_{k-2}\cdot a_k\equiv\dots\equiv a_1\cdo...

- Mon Jun 26, 2017 3:44 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**20947**

### Re: Beginner's Marathon

Solution to problem $24$ ( OVERKILL ) We prove by contradiction. Assume that $n$ divides $a_k(a_1-1)$. Assume that $(a_i, n)=1$. Then $n| a_i(a_{i+1}-1)$ implies $n|a_{i+1}-1$ which implies $a_{i+1}=1$. Then $n|1(a_{i+2}-1)$ which is only possible if $a_{i+2}=1$ but this contradicts the distinction....

- Mon Jun 26, 2017 1:51 am
- Forum: Algebra
- Topic: Good inequality..
- Replies:
**2** - Views:
**4210**

### Re: Good inequality..

By AM-GM inequality, $ac+bc+ab \geq 3(abc)^{2/3}$. Also by AM-GM inequality, $a+b+c \geq 3(abc)^{1/3}$. So, $3 \geq 2+ \frac{3 (abc)^{1/3}}{(a+b+c)}$. Or, $ac+bc+ab \geq 3(abc)^{2/3} \geq (abc)^{2/3} (2+ \frac{3 (abc)^{1/3}}{(a+b+c)})$. Or, $ac+bc+ab \geq 2(abc)^{2/3} + \frac {3abc}{a+b+c}$. Or, $ac...

- Thu Jun 22, 2017 6:51 am
- Forum: Number Theory
- Topic: $x^2 \equiv x (mod n)$
- Replies:
**1** - Views:
**1448**

### Re: $x^2 \equiv x (mod n)$

If we denote $f(n)$ the desired number, then CRT implies $f(mn)=f(m)f(n)$ whenever $(m;n)=1$. So it suffices to find $f(n)$ when $n$ is a prime power. It's quite trivial that there are $2$ solutions for $x $ if $n $ is a prime power. So, the answer is $2^z $ where $z $ is the number of distinct prim...

- Thu Jun 22, 2017 6:20 am
- Forum: Number Theory
- Topic: n^5+n^4
- Replies:
**1** - Views:
**1258**

### Re: n^5+n^4

We rearrange the equation as, $n^5+n^4+1= 7^m $. Observe that $n^5+n^4+1=(n^2+n+1)(n^3-n+1)$. So, each of $n^2+n+1$ and $n^3-n+1$ is $7^x $ for some integer $x \geq 0$. But using the euclidean algorithm, we find that $gcd(n^2+n+1, n^3-n+1)$ divides $7$. So we consider the $4$ equations $n^2+n+1=1$ $...