Bigganchinta is wrong.

- Thu Dec 14, 2017 12:46 pm
- Forum: Higher Secondary Level
- Topic: A strange NUMBER THEORY Problem
- Replies:
**3** - Views:
**53**

Problem 20

A pentagon with all sides equal is given. Prove that the circles having those sides as diameters can't cover the the entire region of that pentagon.

A pentagon with all sides equal is given. Prove that the circles having those sides as diameters can't cover the the entire region of that pentagon.

- Fri Dec 01, 2017 7:30 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**43** - Views:
**2561**

You're actually misinterpreting the question. The problem basically gives us that $(x+1) + (x+2) + ... + (x+y) =976 $. With $x+1$ being the first missing page and $x+y$ the last missing page. So we need to find $y$. $(x+1) + (x+2) + ... + (x+y) = xy + \dfrac {y(y+1)}{2}=976$ So, $y(x+ \dfrac {y+1}{2...

- Mon Nov 27, 2017 4:41 pm
- Forum: Secondary Level
- Topic: Sylhet - 2014
- Replies:
**2** - Views:
**396**

Problem $49$ Let $ABC$ be an acute-angled triangle with $AB\not= AC$. Let $\Gamma$ be the circumcircle, $H$ the orthocentre and $O$ the centre of $\Gamma$. $M$ is the midpoint of $BC$. The line $AM$ meets $\Gamma$ again at $N$ and the circle with diameter $AM$ crosses $\Gamma$ again at $P$. Prove th...

- Sun Nov 26, 2017 7:28 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**109** - Views:
**4809**

Case 1: $a_0\equiv 0\pmod{3}$. We have $a_m\equiv 0\pmod{3}\,\,\forall m\geq 0$. If $a_0=3$ then $a_{3m}=3\,\,\forall m\geq 0$, therefore $a_0=3$ satisfying the condition of the problem. If $a_0=3k$ for some $k>1$. We will prove that there is an index $m_0$ such that $a_{m_0}<a_0$, and therefore (b...

- Sun Oct 01, 2017 10:25 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO $2017$ P$1$
- Replies:
**4** - Views:
**570**

Problem $44$ Let $\triangle ABC$ be an acute angled triangle satisfying the conditions $AB > BC$ and $AC > BC$. Denote by $O$ and $H$ the circumcentre and orthocentre, respectively, of $\triangle ABC$. Suppose that the circumcircle of the triangle $AHC$ intersects the line $AB$ at $M$ diﬀerent from ...

- Fri Sep 01, 2017 2:15 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**109** - Views:
**4809**

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 ...

- Fri Sep 01, 2017 2:00 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**109** - Views:
**4809**

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 tha...

- Wed Jul 19, 2017 12:25 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO $2017$ P$1$
- Replies:
**4** - Views:
**570**

$-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.

- Tue Jul 04, 2017 11:53 pm
- Forum: Number Theory
- Topic: Infinite solutions
- Replies:
**1** - Views:
**216**

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

- Fri Jun 30, 2017 12:31 pm
- Forum: Site Support
- Topic: help please!!!!!!!!
- Replies:
**1** - Views:
**238**