## Search found 181 matches

- Tue Feb 06, 2018 1:14 pm
- Forum: News / Announcements
- Topic: geometry(iranian geometry olympiad 2017)
- Replies:
**4** - Views:
**4195**

### Re: geometry(iranian geometry olympiad 2017)

Okay so you're having difficulty in understanding what clockwise is. In the picture $ABC$ is clockwise and $A_1B_1C_1$ is counterclockwise.

- Sun Feb 04, 2018 2:54 am
- Forum: News / Announcements
- Topic: geometry(iranian geometry olympiad 2017)
- Replies:
**4** - Views:
**4195**

### Re: geometry(iranian geometry olympiad 2017)

What help do you need?

- Sun Feb 04, 2018 2:47 am
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 1
- Replies:
**19** - Views:
**7792**

### Re: BDMO 2017 National round Secondary 1

Both the solutions are completely wrong. Because the 20 cases aren't equally likely.

- Wed Jan 24, 2018 10:09 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 4
- Replies:
**5** - Views:
**2365**

### Re: BDMO 2017 National round Secondary 4

This is a typing mistake, it should be $CM=6.5$.

- Thu Dec 14, 2017 12:46 pm
- Forum: Higher Secondary Level
- Topic: A strange NUMBER THEORY Problem
- Replies:
**4** - Views:
**4854**

### Re: A strange NUMBER THEORY Problem

Bigganchinta is wrong.

- Fri Dec 01, 2017 7:30 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**26489**

### Re: Combi Marathon

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.

- Mon Nov 27, 2017 4:41 pm
- Forum: Secondary Level
- Topic: Sylhet - 2014
- Replies:
**2** - Views:
**1652**

### Re: Sylhet - 2014

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...

- Sun Nov 26, 2017 7:28 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**115** - Views:
**58319**

### Re: Geometry Marathon : Season 3

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 Oct 01, 2017 10:25 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO $2017$ P$1$
- Replies:
**4** - Views:
**6734**

### Re: IMO $2017$ P$1$

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...

- Fri Sep 01, 2017 2:15 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**115** - Views:
**58319**

### Re: Geometry Marathon : Season 3

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 ...