## Search found 11 matches

- Thu May 17, 2018 2:34 pm
- Forum: Primary Level
- Topic: Not too unkind
- Replies:
**2** - Views:
**2843**

### Re: Not too unkind

Solution to a modified G6. Mathmash round 15 P8

- Thu May 17, 2018 2:24 pm
- Forum: Primary Level
- Topic: Not too unkind
- Replies:
**2** - Views:
**2843**

### Not too unkind

. Note that $EF\| ST$ implies that $M$ lies on the median from $A$ of $\triangle {AST}$. As the reflection of $M$ in $ST$ lies on the$\odot{AST}$, we conclude that $AK$ is a symmedian in $\triangle{AST}$, and hence in $\triangle{AEF}$as well. Since $K$ lies on the perpendicular bisector of $EF$, we ...

- Tue Feb 20, 2018 11:32 am
- Forum: Combinatorics
- Topic: triangle in a square has area less than 1/2
- Replies:
**5** - Views:
**5558**

### Re: triangle in a square has area less than 1/2

Take a square of length $n$ and choose $(n+1)^2$ points inside the square. Prove that, you can choose some $3$ points so that the triangle made by them has an area at most $\dfrac12$. It is a nice problem. Sketch of solution: Take the convex hull of points H. Apparently,the hull has area at most n^...

- Mon Feb 19, 2018 1:00 pm
- Forum: Combinatorics
- Topic: triangle in a square has area less than 1/2
- Replies:
**5** - Views:
**5558**

### Re: triangle in a square has area less than 1/2

Silly me.Couldn't latex for shortage of time :"(

- Mon Feb 19, 2018 12:59 pm
- Forum: Combinatorics
- Topic: triangle in a square has area less than 1/2
- Replies:
**5** - Views:
**5558**

### Re: triangle in a square has area less than 1/2

Take a square of length $n$ and choose $(n+1)^2$ points inside the square. Prove that, you can choose some $3$ points so that the triangle made by them has an area at most $\dfrac12$. It is a nice problem. Sketch of solution: Take the convex hull of points . Denote it by H.Apparently,the hull has a...

- Mon Feb 19, 2018 12:45 pm
- Forum: Combinatorics
- Topic: বিয়ে
- Replies:
**6** - Views:
**7897**

### Re: বিয়ে

Dude, it isn't mathoscopeahmedittihad wrote: ↑Sat Feb 17, 2018 1:41 amYes, yes I am proud of my knowledge of marriage procedures.SN.Pushpita wrote: ↑Thu Feb 15, 2018 1:38 pmAre you proud of your knowledge regarding Hall's Marriage Lemma?:p

- Thu Feb 15, 2018 2:24 pm
- Forum: Combinatorics
- Topic: n-series
- Replies:
**1** - Views:
**4094**

### Re: n-series

First observe that for any sequence, if the first 3 terms are a,b and c respectively,then either b<a <c or c <a <b will hold. Now repeating the condition given in the question and fixing the number of terms u can get the order of each sequence.So FOR each subset of cardinality >=3 of {1,2,3,....n} u...

- Thu Feb 15, 2018 1:38 pm
- Forum: Combinatorics
- Topic: বিয়ে
- Replies:
**6** - Views:
**7897**

### Re: বিয়ে

Are you proud of your knowledge regarding Hall's Marriage Lemma?:p

- Wed Feb 14, 2018 1:20 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**24851**

### Re: Combi Marathon

First one has holes but the second one is correct.

- Wed Feb 14, 2018 1:06 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**24851**

### 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. Sketch of solution: This claim is true for convex pentagon.For any convex pentagon,the summation of internal angles is 540 and 540-3×60=2×1...