Search found 136 matches
- Wed Jan 14, 2015 2:44 am
- Forum: Combinatorics
- Topic: n+1 rows and columns
- Replies: 10
- Views: 8587
Re: n+1 rows and columns
Call the black coloring method described in the problem a good coloring, and call the set of cells described in the problem (no $2$ on same row/column) a good set. Lemma: A good coloring is possible iff at least one row/column is completely black. Proof: Suppose not, i.e. no row or column of the boa...
- Fri Dec 19, 2014 12:02 am
- Forum: Algebra
- Topic: Existence of Multiplicative Inverse in a Commutative Ring
- Replies: 2
- Views: 2822
Re: Existence of Multiplicative Inverse in a Commutative Rin
A Quicker Solution: Define the followings \[\sum_{j=0}^{n-1} q_j\alpha^j=\gamma\in\mathbb{A},~~P(x)=\sum_{j=0}^n a_jx^j,~~\Gamma(x)=\sum_{j=0}^{n-1}q_jx^j.\] Since $P$ is irreducible and $\deg P=1+\deg \Gamma$ it follows that $P$ and $\Gamma$ are coprime. Now by the extended Euclidean algorithm for ...
- Thu Dec 18, 2014 12:06 am
- Forum: Algebra
- Topic: Existence of Multiplicative Inverse in a Commutative Ring
- Replies: 2
- Views: 2822
Existence of Multiplicative Inverse in a Commutative Ring
Let $\alpha$ be an algebraic number. Prove that the Euclidean domain $\mathbb{Q}[\alpha]$ is in fact a field. In other words, let $\alpha$ be a root of the polynomial $P$ such that $P$ is irreducible over $\mathbb Q$. Let $n=\deg P$. Define $\mathbb{A}=\left\{\displaystyle\sum_{j=0}^{n-1} q_j\alpha^...
- Sat Dec 13, 2014 10:49 pm
- Forum: Combinatorics
- Topic: Coins on a polygon
- Replies: 1
- Views: 2473
Re: Coins on a polygon
Simply assigning the weight $j$ to the coins at $A_j$ reveals $n=6x\pm 1$.
- Mon Dec 08, 2014 12:51 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 134259
Re: IMO Marathon
Solution $\boxed{39}$ Let the incircle $\Gamma$ of $\triangle ABC$ touch sides $a,b,c$ at $P,Q,R$ respectively. Let $\omega_A$ touch $\Gamma$ at $D$. Also let $T_A$ be the midpoint of arc $BC$ not containing $D$. It then easily follows that $\triangle T_ABP\sim \triangle T_ADB$ so $T_AB^2=T_AD\cdot...
- Sat Nov 29, 2014 10:12 pm
- Forum: Geometry
- Topic: powers are equal
- Replies: 3
- Views: 3612
Re: powers are equal
Sorry I messed the construction up.
- Wed Nov 26, 2014 1:21 pm
- Forum: Geometry
- Topic: powers are equal
- Replies: 3
- Views: 3612
Re: powers are equal
I might be wrong, but $A,B,C$ all lie on the circles they are meant to be respected to. So they have equal power $0$.
- Tue Nov 18, 2014 11:33 pm
- Forum: Combinatorics
- Topic: A Beautiful Combi
- Replies: 2
- Views: 3128
Re: A Beautiful Combi
My solution relies on a weighting invariance. Assign the stone at $(x,y)$ the weight $\dfrac{1}{2^{x+y}}$. The reason for this choice is that under any valid move, the total weight remains constant, and that is $1$. So at some point in the process, if we exclude the points $(a,b)$ with $a+b\le 3$ th...
- Mon Nov 17, 2014 2:23 pm
- Forum: Number Theory
- Topic: Numbers from blackboard
- Replies: 3
- Views: 3294
Re: Numbers from blackboard
Suppose you started with two positive integers in the blackboard and completed the game with finitely many steps. Prove you get two equal positive integers at last . Maybe a typo? Cause, the problem is obvious otherwise. (According to your definitions, the game would not end unless arrived to two e...
- Tue Nov 11, 2014 1:29 am
- Forum: Combinatorics
- Topic: A Beautiful Combi
- Replies: 2
- Views: 3128
A Beautiful Combi
Suppose we have a collection of infinite stones. On the Cartesian plane, a stone is kept on $(0,0)$. In a move, we can remove a stone from $(m,n)$ (provided that it contains a stone) and keep two stones on $(m+1,n)$ and $(m,n+1)$ (provided that these two points don't have any stones). Prove that at ...