## Search found 153 matches

Fri Dec 02, 2011 10:09 pm
Forum: Junior Level
Topic: Math problem 2 (need solution)
Replies: 9
Views: 3502

### Re: Math problem 2 (need solution)

I misread as it asked the difference between maximum and minimum perimeter. Actually $2\left(\max\{AB,BC,CA\}-\min\{AB,BC,CA\}\right)$ doesn't express an area here!
Fri Dec 02, 2011 2:22 pm
Forum: Junior Level
Topic: Math problem 2 (need solution)
Replies: 9
Views: 3502

### Re: Math problem 2 (need solution)

Not actually, you can draw only 3 parallelogram with 3 given vertexes. So if the given points are $A, B,C$ then the answer of your problem is $2\left(\max\{AB,BC,CA\}-\min\{AB,BC,CA\}\right)$ By the way, if you want to post something drawn, you can draw it in your PC and then submit it as a attachme...
Fri Dec 02, 2011 12:51 pm
Forum: Junior Level
Topic: Math problem 2 (need solution)
Replies: 9
Views: 3502

### Re: Math problem 2 (need solution)

ataher.sams wrote:One can draw as many parallelograms as possible keeping those three points as the three vertices of the parallelogram
How ?
Fri Nov 25, 2011 3:21 pm
Forum: Number Theory
Topic: Hard sequence
Replies: 2
Views: 1232

Nice problem! :) $[0]=(1,1)$ $[1]=(1,2,1)$ $[2]=(1,3,2,3,1)$ . . . $[n]=(a_1, a_2, a_3, \cdots , a_m)$ $S(n) = a_1^3 + a_2^3 + \cdots + a_m^3$ $T(n) = a_1 a_2 (a_1+a_2) + a_2 a_3 (a_2+a_3) + \cdots + a_{m-1}a_m ( a_{m-1}+a_m )$ $c_n = S(n)+T(n) - 1$ Note that, $[n+1]=(a_1, a_1+a_2,a_2,a_2+a_3,\cdots... Fri Nov 25, 2011 2:08 pm Forum: Number Theory Topic: Find all integers Replies: 3 Views: 1543 ### Re: Find all integers "from which every point lies outside the circle which contains all the other points and has it's centre in one of it.", I can't get it. Tue Nov 22, 2011 2:34 am Forum: Algebra Topic: Functional Equation [Own] Replies: 2 Views: 1315 ### Re: Functional Equation [Own] Complete Solution, Sat Nov 19, 2011 12:45 pm Forum: Algebra Topic: Functional Equation [Own] Replies: 2 Views: 1315 ### Re: Functional Equation [Own] At last solved :mrgreen: Complete solutions:$ f(x) =\begin{cases}c &\mbox{if }x\geq a\\ a+h(x) &\mbox{if }a\geq x\geq\frac{r}{2a}\\ \frac{r}{2x}+h(x) &\mbox{if }\frac{r}{2a}\geq x\end{cases} $Where$ 2c^{2}\geq 2ac\geq r\geq a $and$ h:(0,a]\longrightarrow [0,\infty) $is any continuous function ... Sat Nov 05, 2011 11:30 pm Forum: National Math Camp Topic: solutions to camp exam problem Replies: 28 Views: 8275 ### Re: solutions to camp exam problem হুম, আমি চাইসিলাম ডানপক্ষে যেন$|x-y|$থাকে, তাই$(x,\frac{x+y}{2} )$নিসি, যেহেতু মূল সমীকরণ x আর y এর জন্য সমরূপ, তাই,$(y ,\frac{x+y}{2} )$-ও নিসি এবং শেষে বাকীপদগুলা নিচিহ্ন করার জন্য$(\frac{3x+y}{4}, \frac{x+3y}{4}$নিসি। Sat Nov 05, 2011 8:17 pm Forum: National Math Camp Topic: solutions to camp exam problem Replies: 28 Views: 8275 ### Re: solutions to camp exam problem @Joty: P2: আমরা জানি$|a| + |b| \geq |a+b|$এবং আরো জানি,$f'(x) \geq 0$হলে$f$increasing হবে। P9: এটা আমি পুরা লিখতে ভুলে গেসি। খেয়াল কর,$f(x) = \frac{1}{\sqrt(1+e^{2x})}\$ কনভেক্স। তারপর Jensen মার। P10: জানি না :-/ তবে এটা মোটামুটি এখন Intuition হয়ে গেছে, অনেক বেশি Function equation করার ফল। ...
Sat Nov 05, 2011 8:10 pm
Forum: National Math Camp
Topic: solutions to camp exam problem
Replies: 28
Views: 8275

### Re: solutions to camp exam problem

Texmacs :
sudo apt-get install texmacs