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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.
by ahmedittihad
Fri Dec 01, 2017 7:30 pm
 
Forum: Combinatorics
Topic: Combi Marathon
Replies: 43
Views: 2556

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

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...
by ahmedittihad
Sun Nov 26, 2017 7:28 pm
 
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 108
Views: 4745

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...
by ahmedittihad
Sun Oct 01, 2017 10:25 pm
 
Forum: International Mathematical Olympiad (IMO)
Topic: IMO $2017$ P$1$
Replies: 4
Views: 561

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 ...
by ahmedittihad
Fri Sep 01, 2017 2:15 pm
 
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 108
Views: 4745

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 ...
by ahmedittihad
Fri Sep 01, 2017 2:00 pm
 
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 108
Views: 4745

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 tha...
by ahmedittihad
Wed Jul 19, 2017 12:25 am
 
Forum: International Mathematical Olympiad (IMO)
Topic: IMO $2017$ P$1$
Replies: 4
Views: 561

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.
by ahmedittihad
Tue Jul 04, 2017 11:53 pm
 
Forum: Number Theory
Topic: Infinite solutions
Replies: 1
Views: 215

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

You upload attachments by clicking the 'Upload Attachment' and choosing a file to upload.
by ahmedittihad
Fri Jun 30, 2017 12:31 pm
 
Forum: Site Support
Topic: help please!!!!!!!!
Replies: 1
Views: 236

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 $A_1$, $B_1$ ...
by ahmedittihad
Mon Jun 26, 2017 4:04 am
 
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 68
Views: 3727
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