Problem 8:

Suppose that 5 points lie on a sphere. Prove that there exists a closed semi-sphere (half a sphere including boundary), which contains 4 of the points.

Problem 8:

Suppose that 5 points lie on a sphere. Prove that there exists a closed semi-sphere (half a sphere including boundary), which contains 4 of the points.

Suppose that 5 points lie on a sphere. Prove that there exists a closed semi-sphere (half a sphere including boundary), which contains 4 of the points.

- Thu Mar 30, 2017 1:21 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**22** - Views:
**380**

P7. There are 2000 points on a circle and each point is given a number which is equal to the average of the two numbers which are its nearest neighbors. Show that all the numbers must be equal. An alternate Solution: We use extremal principal. Since there a finite number of numbers, there must exis...

- Thu Mar 30, 2017 1:15 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**22** - Views:
**380**

Given $2n$ points in the plane with no three collinear, show that it is possible to pair them up in such a way that the $n$ line segments joining paired points do not intersect. Here is my take. Take all possible pairings of the points, and select the one with minimum sum of the length of the segme...

- Mon Mar 06, 2017 6:13 pm
- Forum: Junior Level
- Topic: Don't Intersect Please :D
- Replies:
**7** - Views:
**167**

$\text{Problem 14:}$ There are $ n + 1$ cells in a row labeled from $ 0$ to $ n$ and $ n + 1$ cards labeled from $ 0$ to $ n$. The cards are arbitrarily placed in the cells, one per cell. The objective is to get card $ i$ into cell $ i$ for each $ i$. The allowed move is to find the smallest $ h$ su...

- Wed Mar 01, 2017 5:44 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**38** - Views:
**783**

$\text{Solution to Problem 13}$ Lemma: Let $S$ be a set of $m$ integers. Let $T$ be the set of integers that can be written as a sum of $t$ distinct elements of $S$. Then $|T|\ge t(m-t)+1$. Proof: Let $a_1<a_2<a_3<\ldots <a_m$ be the elements of $S$. Now, take the sum $a_1+a_2+\ldots +a_t$. Now, we ...

- Wed Mar 01, 2017 5:25 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**38** - Views:
**783**

Here is an alternate solution to problem 36 We apply cartesian coordinates. fig.png Note that the problem is equivalent to proving $F',D,E$ collinear in the diagram. Let $U\equiv (a,1), V\equiv (-a,1), T\equiv (at,t), S\equiv (-as,s)$. Now, let $E \equiv (p,q)$. We get, $\dfrac{q-t}{p-at}=-a$ and $...

- Tue Feb 28, 2017 2:32 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**81** - Views:
**1657**

$\text{Problem 12}$

On a circle there are $150$ red and $150$ blue arcs given in such a way that each red arc intersects each blue one. Prove that there exists a point contained by at least $150$ of the given colored arcs.

On a circle there are $150$ red and $150$ blue arcs given in such a way that each red arc intersects each blue one. Prove that there exists a point contained by at least $150$ of the given colored arcs.

- Mon Feb 27, 2017 11:26 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**38** - Views:
**783**

$\text{Solution to problem 11}$ Translate the whole sequence by $-m$. That way, the nine terms sum to $0$ and the triples that sum to a non-negative number contribute to $A$. Now, take all the $\dbinom{9}{3}=84$ triples. We can arrange them into sets of $3$ triples so that the union of the triples c...

- Mon Feb 27, 2017 11:05 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**38** - Views:
**783**

$\text{Problem 10:}$ 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 block...

- Mon Feb 27, 2017 12:10 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**38** - Views:
**783**

$\text{Solution to problem 9:}$ The answer is $0+1+2+\ldots +(n-1)=\dfrac{n(n-1)}{2}$. Label the $i^{th}$ person as $p_i$ and let the number of coins with him be $c_i$. If $c_i=i-1$, then after each minute, the number of coins are simply permuted and thus it can go forever. Now, if a person can give...

- Mon Feb 27, 2017 12:00 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**38** - Views:
**783**