## Search found 147 matches

### Re: A strange NUMBER THEORY Problem

Bigganchinta is wrong.
Thu Dec 14, 2017 12:46 pm

Forum: Higher Secondary Level
Topic: A strange NUMBER THEORY Problem
Replies: 3
Views: 61

### 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.
Fri Dec 01, 2017 7:30 pm

Forum: Combinatorics
Topic: Combi Marathon
Replies: 43
Views: 2566

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: 397 ### 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 Nov 26, 2017 7:28 pm Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 109 Views: 4825 ### 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: 581 ### 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: 109 Views: 4825 ### 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 ...
Fri Sep 01, 2017 2:00 pm

Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 109
Views: 4825

### 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

Topic: IMO $2017$ P$1$
Replies: 4
Views: 581

### 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: 219