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:
**5** - Views:
**152**

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:
**73**

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:
**165**

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:
**165**

$f^3(1) + f^3(2) + ... + f^3(n) = (\frac{f(n)f(n+1)}{2})^2...(i)$ $f^3(1) + f^3(2) + ... + f^3(n) + f^3(n+1)= (\frac{f(n+1)f(n+2)}{2})^2 ...(ii)$ $(ii) - (i) \Rightarrow f(n+1) = \frac{f^2(n+2)}{4} - \frac{f^2(n)}{4} = (\frac{f(n+2) + f(n)}{2})(\frac{f(n+2) - f(n)}{2})$ 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:
**120**

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:
**120**

Let , $\angle CPY = \angle CBY = \angle p$ $\angle PBC = \angle PYC = \angle q$ $CX = CY \Rightarrow \angle CYX = \angle CXY = \angle q$ $AB = CY , BY = BY , \angle BAY = \angle BCY$ So,$\triangle ABY \cong \triangle BCY$ Thus, $\angle CBY = \angle AYB = \angle ACB = \angle p$ From $\triangle PXC$, ...

- Thu Feb 23, 2017 1:11 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/2
- Replies:
**5** - Views:
**120**

Thanks to @Thanic nur samin .Can anyone give a synthetic solution?

- Mon Feb 20, 2017 2:46 pm
- Forum: Geometry
- Topic: Looking for non-trig solution
- Replies:
**3** - Views:
**67**