Solution to problem $\boxed{10}$: We partition the square in $4$ squares with the length of the side $1$.According to PHP, there is a square which contains at least $3$ of the points .We can see that the area of this triangle is smaller than the area of a triangle with the vertices on the sides of t...

- Sun Apr 09, 2017 4:45 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**2802**

Problem $\boxed{6}$ Let $ABC$ be an acute triangle. The lines $l_1$ and $l_2$ are perpendicular to $AB$ at the points $A$ and $B$, respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$ intersect $l_1$ and $l_2$ at the points $E$ and $F$, respectively. If $D$...

- Tue Mar 28, 2017 10:53 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**2802**

Solution of Problem $\boxed{2}$: $\frac{AK}{AN} = \frac{1}{2} = \cos 60$ Thus,$\triangle AKN$ is a $30-60-90$ triangle. Let,$\angle B = b$.So,$\angle C = 120-b$ By,some angle chasing , we will get $\angle KMN =180 - (\angle KMB + \angle NMC)$ $180 - (\angle KBM + \angle NCM) = 180 - (b+120-b)$ $=60$...

- Fri Mar 24, 2017 1:28 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**2802**

A simpler solution: Draw two diagonals $AC$ and $BD$ and let them intersect in $O$. $AO = BO = CO = DO$ Thus,in $\triangle AEC$ , $EO$ is the median. Appolonius' theorem implies that $AE^2 + CE^2 = 2(AO^2 + EO^2)$ Similarly , from $\triangle BED$ , $DE^2 + BE^2 = 2(BO^2 + EO^2) = 2(AO^2 + EO^2)$ Thu...

- Wed Mar 08, 2017 10:59 pm
- Forum: Divisional Math Olympiad
- Topic: BDMO REGIONAL 2015
- Replies:
**2** - Views:
**167**

I think 70% or more are introduced to convex hull by this 'rubber-band' concept.

- Mon Mar 06, 2017 12:03 pm
- Forum: Junior Level
- Topic: Don't Intersect Please :D
- Replies:
**7** - Views:
**387**

Actually,@thamimzahin,the idea which you have is the main concept of convex hull.

http://mathworld.wolfram.com/ConvexHull.html

http://mathworld.wolfram.com/ConvexHull.html

- Sun Mar 05, 2017 10:41 pm
- Forum: Junior Level
- Topic: Don't Intersect Please :D
- Replies:
**7** - Views:
**387**

$f^3(1) + f^3(2) + ... + f^3(n) = (\frac{f(n)f(n+1)}{2})^2...(i)$ $f^3(1) + f^3(2) + ... + f^3(n) + f^3(n+1)= (\frac{f(n+1)f(n+2)}{2})^2 ...(ii)$ $(ii) - (i) \Rightarrow f(n+1) = \frac{f^2(n+2)}{4} - \frac{f^2(n)}{4} = (\frac{f(n+2) + f(n)}{2})(\frac{f(n+2) - f(n)}{2})$ Since, $f(n+1) > 0 \Rightarro...

Find all function $f: N \rightarrow N$ such that ,

$f^3(1) + f^3(2) + ... + f^3(n) = (\frac{f(n)f(n+1)}{2})^2$ and $f(1) = 1$

$f^3(1) + f^3(2) + ... + f^3(n) = (\frac{f(n)f(n+1)}{2})^2$ and $f(1) = 1$

How we can write $\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$?

- Sun Feb 26, 2017 3:21 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/2
- Replies:
**5** - Views:
**251**

Here is the official solution :

We know that $CX = CY$ therefore:

$\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$ [But how?]

Also we have $AB = CY$ therefore $\widehat{AP} + \widehat{CY} = \widehat{AP} + \widehat{AB} = \widehat{PB}$

So $PB = PC$.

We know that $CX = CY$ therefore:

$\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$ [But how?]

Also we have $AB = CY$ therefore $\widehat{AP} + \widehat{CY} = \widehat{AP} + \widehat{AB} = \widehat{PB}$

So $PB = PC$.

- Thu Feb 23, 2017 1:21 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/2
- Replies:
**5** - Views:
**251**