## Search found 143 matches

### 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...
Sun Oct 01, 2017 10:25 pm

Forum: International Mathematical Olympiad (IMO)
Topic: IMO $2017$ P$1$
Replies: 4
Views: 368

### 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 ...
Fri Sep 01, 2017 2:15 pm

Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 92
Views: 3914

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: 92 Views: 3914 ### 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... Wed Jul 19, 2017 12:25 am Forum: International Mathematical Olympiad (IMO) Topic: IMO$2017$P$1$Replies: 4 Views: 368 ### 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. Tue Jul 04, 2017 11:53 pm Forum: Number Theory Topic: Infinite solutions Replies: 1 Views: 180 ### Re: help please!!!!!!!! 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: 164 ### 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 4:04 am Forum: Junior Level Topic: Beginner's Marathon Replies: 68 Views: 3333 ### 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 3:58 am

Forum: Junior Level
Topic: Beginner's Marathon
Replies: 68
Views: 3333

### 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 3:44 am

Forum: Junior Level
Topic: Beginner's Marathon
Replies: 68
Views: 3333

### 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...
Mon Jun 26, 2017 1:51 am

Forum: Algebra
Topic: Good inequality..
Replies: 2
Views: 218
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