## Let us help one another preparing for BdMO national 2013

nafistiham
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### Let us help one another preparing for BdMO national 2013

Divisionals are over. So, around a thousand participants are preparing for national. This topic is just as its name. I request interested people to join with me.

I plan to post problems which are just our national level. neither IMO nor divisional. I believe the practice for contest works best of it is in the contest range. So, please post no problem which are too hard or too easy.

Personally, I am going to post which I can't solve yet. But, I hope people will share what they can do with hidden solutions.

Just, do one favour. Post problems with category types. I mean, tag a category name with the problem. Just as you think it can be. Because, before solving one may not know how easy or hard it is.

If, this topic goes on, I think the participants who are first or second to attend BdMO and shy to post their problems here (believe me, there is a lot of people like this I know) will also be helpful by just going through the post.

Moreover, if it turns out that, only I am the problem poster, I'll make arrangements to edit the topic name as "Help me preparing for BdMO national 2013"

$1.$ Pentagon equlibrium.
Secondary, Higher Secondary (and Junior may be)
from Plane Euclidean Geometry (p-98)
it is beautiful
Regular pentagon.png (33.61 KiB) Viewed 4991 times
$AB=BC=CD=DE=EA$
$F$ is an arbitary point in $AE$ arc of the circumcircle.
prove,
$FA+FC+FE=FB+FD$
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

*Mahi*
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### Re: Let us help one another preparing for BdMO national 2013

nafistiham wrote: $1.$ Pentagon equlibrium.
Secondary, Higher Secondary (and Junior may be)
from Plane Euclidean Geometry (p-98)
Regular pentagon.png
$AB=BC=CD=DE=EA$
$F$ is an arbitary point in $AE$ arc of the circumcircle.
prove,
$FA+FC+FE=FB+FD$
Also, I appreciate the effort.

Use $L^AT_EX$, It makes our work a lot easier!

sakib.creza
Posts: 26
Joined: Sat Nov 03, 2012 6:36 am

### Re: Let us help one another preparing for BdMO national 2013

Tiham Bhai, I really find it a timely and helpful initiative.
Here's a problem I couldn't solve.
It's for Secondary and Higher Secondary.
Source- BDMO 2007
2. If $x_1$ , $x_2$ are the zeros of the polynomial $x^2 - 6x +1$, then prove that for every
nonnegative integer $n$ , $x_1^n +x_2^n$ is an integer and not divisible by $5$ .

Posts: 244
Joined: Sat May 07, 2011 12:36 pm
Location: B.A.R.D , kotbari , Comilla

### Re: Let us help one another preparing for BdMO national 2013

($x_1=a=3+\sqrt{8},y_1=b=3-\sqrt{8}$)

Now Using vieta we have $(a+b)=6,ab=1$

$a^n+b^n=(3+\sqrt{8})^n+(3-\sqrt{8})^n$
We now that $r+1$th term of the expantion of $(3+\sqrt{8})^n$ is $\displaystyle \binom {n}{r}3^{n-r}.(\sqrt{8})^r$ , and for $(3-\sqrt{8})^n$ it is $\displaystyle \binom {n}{r}3^{n-r}.(-\sqrt{8})^r$

If $r$ is even , then both $\displaystyle \binom {n}{r}3^{n-r}.(\sqrt{8})^r,\binom {n}{r}3^{n-r}.(-\sqrt{8})^r$ are integer . If $r$ is odd they will have a same value but different sign ,thus makes 0 after katakati .

So $a^n+b^n=(3+\sqrt{8})^n+(3-\sqrt{8})^n$ is an integer for all $n$.

Now note the identity : $a^n+b^n=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})=6(a^{n-1}+b^{n-1})-(a^{n-2}+b^{n-2})$

So $a^n+b^n \equiv 6(a^{n-1}+b^{n-1})-(a^{n-2}+b^{n-2}) \equiv (a^{n-1}+b^{n-1})-(a^{n-2}+b^{n-2}) (mod5)$
Now we may use induction .
$\frac{1}{0}$

Fahim Shahriar
Posts: 138
Joined: Sun Dec 18, 2011 12:53 pm

### Re: Let us help one another preparing for BdMO national 2013

$x_1 = 3 + 2 \sqrt {2}$ & $x_2 = 3 - 2 \sqrt {2}$

$(3 + 2 \sqrt {2})^n + (3 - 2 \sqrt {2})^n$

$(1 + \sqrt {2})^{2n} + (1 - \sqrt {2})^{2n}$

$2*(1 + ^{2n} C_2 \times 2^2 + ^{2n} C_4 \times 2^4 +.. .. ..+ 2^n)$

We will not get any irrational number $[ \sqrt {2} ]$ here because they are getting cancelled by each other.
And it doesn't have factor 5. Hence it's not divisible by 5.
Last edited by *Mahi* on Fri Feb 01, 2013 9:25 pm, edited 1 time in total.
Reason: LaTeXed properly
Name: Fahim Shahriar Shakkhor
Notre Dame College

Posts: 244
Joined: Sat May 07, 2011 12:36 pm
Location: B.A.R.D , kotbari , Comilla

### Re: Let us help one another preparing for BdMO national 2013

Fahim Shahriar wrote:
And it doesn't have factor 5. Hence it's not divisible by 5.
How come ?
$\frac{1}{0}$

Fahim Shahriar
Posts: 138
Joined: Sun Dec 18, 2011 12:53 pm

### Re: Let us help one another preparing for BdMO national 2013

I only calculated on my mind-
When I expand it, $1+ ^{2n} C_2 * 2^2 + ^{2n} C_2 * 2^4 +..$
Here every $2$ has even number as power. Which means it is either $(1or-1) \equiv (mod 5)$
Then I had a look on Pascal's triangle. The sum of the coefficients is $2^{n-1}$.
Here my problem starts. While doing it on my mind taking n once even & once odd, I was done.But after ur reply, I sat with pen & paper & tried again. This time I am stuck.
My mob cant read LTX without bitmap, cant read ur full soluti
Name: Fahim Shahriar Shakkhor
Notre Dame College

sakib.creza
Posts: 26
Joined: Sat Nov 03, 2012 6:36 am

### Re: Let us help one another preparing for BdMO national 2013

$a^n+b^n \equiv 6(a^{n-1}+b^{n-1})-(a^{n-2}+b^{n-2}) \equiv (a^{n-1}+b^{n-1})-(a^{n-2}+b^{n-2}) (mod5)$
Now we may use induction .

*Mahi*
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### Re: Let us help one another preparing for BdMO national 2013

This follows from the line above that, $a^n+b^n=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})=6(a^{n-1}+b^{n-1})-(a^{n-2}+b^{n-2})$ with $a+b = 6$ and $ab =1$

Use $L^AT_EX$, It makes our work a lot easier!

$a^n+b^n \equiv 6(a^{n-1}+b^{n-1})-(a^{n-2}+b^{n-2}) \equiv (a^{n-1}+b^{n-1})-(a^{n-2}+b^{n-2}) (mod5)$