It isn't Octahedron" at all! It is "Octagon".
This is octahedron :
Search found 65 matches
- Fri Feb 02, 2018 10:41 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 6
- Replies: 9
- Views: 8516
- Wed Jan 24, 2018 11:09 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 4
- Replies: 5
- Views: 4683
Re: BDMO 2017 National round Secondary 4
How can it be $AD \perp BC$ and $\angle ADC = 90^o$ ? Is it a valid condition?
- Sun Apr 09, 2017 4:45 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies: 68
- Views: 47404
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...
- Tue Mar 28, 2017 10:53 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies: 68
- Views: 47404
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 $D$...
- Fri Mar 24, 2017 1:28 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies: 68
- Views: 47404
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 $\angle KMN =180 - (\angle KMB + \angle NMC)$ $180 - (\angle KBM + \angle NCM) = 180 - (b+120-b)$ $=60$...
- Wed Mar 08, 2017 10:59 pm
- Forum: Divisional Math Olympiad
- Topic: BDMO REGIONAL 2015
- Replies: 3
- Views: 4140
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...
- Mon Mar 06, 2017 12:03 pm
- Forum: Junior Level
- Topic: Don't Intersect Please :D
- Replies: 7
- Views: 5596
Re: Don't Intersect Please :D
I think 70% or more are introduced to convex hull by this 'rubber-band' concept.
- Sun Mar 05, 2017 10:41 pm
- Forum: Junior Level
- Topic: Don't Intersect Please :D
- Replies: 7
- Views: 5596
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
http://mathworld.wolfram.com/ConvexHull.html
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...
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 $f(1) = 1$
$f^3(1) + f^3(2) + ... + f^3(n) = (\frac{f(n)f(n+1)}{2})^2$ and $f(1) = 1$