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Re: Beginner's Marathon

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...
by Absur Khan Siam
Sun Apr 09, 2017 4:45 pm
 
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 63
Views: 2123

Re: Beginner's Marathon

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 $D$...
by Absur Khan Siam
Tue Mar 28, 2017 10:53 pm
 
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 63
Views: 2123

Re: Beginner's Marathon

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 $\angle KMN =180 - (\angle KMB + \angle NMC)$ $180 - (\angle KBM + \angle NCM) = 180 - (b+120-b)$ $=60$...
by Absur Khan Siam
Fri Mar 24, 2017 1:28 am
 
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 63
Views: 2123

Re: BDMO REGIONAL 2015

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...
by Absur Khan Siam
Wed Mar 08, 2017 10:59 pm
 
Forum: Divisional Math Olympiad
Topic: BDMO REGIONAL 2015
Replies: 2
Views: 144

Re: Don't Intersect Please :D

I think 70% or more are introduced to convex hull by this 'rubber-band' concept.:D
by Absur Khan Siam
Mon Mar 06, 2017 12:03 pm
 
Forum: Junior Level
Topic: Don't Intersect Please :D
Replies: 7
Views: 312

Re: Don't Intersect Please :D

Actually,@thamimzahin,the idea which you have is the main concept of convex hull.
http://mathworld.wolfram.com/ConvexHull.html
by Absur Khan Siam
Sun Mar 05, 2017 10:41 pm
 
Forum: Junior Level
Topic: Don't Intersect Please :D
Replies: 7
Views: 312

Re: Easy FE

$f^3(1) + f^3(2) + ... + f^3(n) = (\frac{f(n)f(n+1)}{2})^2...(i)$ $f^3(1) + f^3(2) + ... + f^3(n) + f^3(n+1)= (\frac{f(n+1)f(n+2)}{2})^2 ...(ii)$ $(ii) - (i) \Rightarrow f(n+1) = \frac{f^2(n+2)}{4} - \frac{f^2(n)}{4} = (\frac{f(n+2) + f(n)}{2})(\frac{f(n+2) - f(n)}{2})$ Since, $f(n+1) > 0 \Rightarro...
by Absur Khan Siam
Tue Feb 28, 2017 12:37 pm
 
Forum: Algebra
Topic: Easy FE
Replies: 1
Views: 85

Easy FE

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 $f(1) = 1$
by Absur Khan Siam
Tue Feb 28, 2017 12:22 pm
 
Forum: Algebra
Topic: Easy FE
Replies: 1
Views: 85

Re: IGO 2016 Elementary/2

How we can write $\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$?
by Absur Khan Siam
Sun Feb 26, 2017 3:21 pm
 
Forum: Geometry
Topic: IGO 2016 Elementary/2
Replies: 5
Views: 210

Re: IGO 2016 Elementary/2

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 $PB = PC$.
by Absur Khan Siam
Thu Feb 23, 2017 1:21 pm
 
Forum: Geometry
Topic: IGO 2016 Elementary/2
Replies: 5
Views: 210
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