Search found 110 matches
- Tue Feb 28, 2017 1:01 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies: 48
- Views: 44351
Re: Combi Marathon
$\text{Solution } 12$ Below, we distinguish between arc and coloured arc by defining an arc to be any sector of the circumference of the circle. So all coloured arcs are arcs, but not vice versa. Assume that there is no point that belongs to $n$ coloured arcs. The way we progress is by noting that i...
- Mon Feb 27, 2017 5:26 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies: 48
- Views: 44351
Re: Combi Marathon
By general consensus among most participants in this marathon, we won't be posting a solution of the above well-known problem. You can check for solutions here . $\text{Problem 11}$ Let $a_1,a_2,...,a_9$ be nine real numbers, not necessarily distinct, with average $m$. Let $A$ denote the number of t...
- Mon Feb 27, 2017 12:36 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies: 48
- Views: 44351
Re: Combi Marathon
$\text{Problem 10:}$ Consider $2009$ cards, each having one gold side and one black side, lying in parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a bloc...
- Sat Feb 25, 2017 12:46 am
- Forum: Social Lounge
- Topic: সেঞ্চুরি ! সেঞ্চুরি !! সেঞ্চুরি !!!
- Replies: 10
- Views: 14113
Re: সেঞ্চুরি ! সেঞ্চুরি !! সেঞ্চুরি !!!
Congrats on the thousandth post (am posting it here to get my own post-count closer to hundred )
- Tue Feb 21, 2017 12:35 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies: 48
- Views: 44351
Re: Combi Marathon
$\text{Problem } 5$ Let $m, n$ be positive integers with $m > 1$. Anastasia partitions the integers $1, 2, \dots , 2m$ into $m$ pairs. Boris then chooses one integer from each pair and finds the sum of these chosen integers. Prove that Anastasia can select the pairs so that Boris cannot make his sum...
- Tue Feb 21, 2017 12:32 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies: 48
- Views: 44351
Re: Combi Marathon
Problem 4: Each edge of a polyhedron is oriented with an arrow such that at each vertex, there is at least on arrow leaving the vertex and at least one arrow entering the vertex. Does there always exists two faces on the polyhedron such that the edges on each of it's boundary form a directed cycle?...
- Mon Feb 20, 2017 10:35 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies: 48
- Views: 44351
Re: Combi Marathon
Problem $3$ Let $n$ be a positive integer. At each of the $2n$ points around a circle we place discs with one white side and one black side. We may perform the following move: select a black disc and flip over its two neighbors. Find all initial configurations from which some sequence of moves lead...
- Mon Feb 20, 2017 10:17 pm
- Forum: Geometry
- Topic: A Problem for Dadu
- Replies: 2
- Views: 3080
Re: A Problem for Dadu
We will show that the four lines are concurrent at their midpoints. Lemma : $AH_AH_BB$ is a parallelogram Proof : Since $AH_A=2\times \text{distance of O to CD}=BH_B$, and $AH_A||BH_B$, we are done. Now it easily follows that the diagonals $AH_B$ and $BH_A$ bisect each other. We can do this for ever...
- Mon Feb 20, 2017 2:26 pm
- Forum: Combinatorics
- Topic: Combi Solution Writing Threadie
- Replies: 10
- Views: 13299
Re: Combi Solution Writing Threadie
Nice solu Tanmoy. However, the things that you are doing seem magic until we get to the end. Perhaps you should add a brief summary at the beginning. Something like "We will show that every set of four points can contribute at most $2$ to the original sum".
- Sun Feb 19, 2017 11:23 am
- Forum: Algebra
- Topic: Instructive FE (I desperately need that topic list)
- Replies: 1
- Views: 2176
Re: Instructive FE (I desperately need that topic list)
Some hints. First try to guess the answer. Done? Let's move on. Now there aren't a lot of things you can do in this problem apart from proving $f$ is injective, and even that doesn't yield much. So you can only plug in values. Plug in $n=1$. First assume that $f(1)=1$. Keep reiterating the values to...