Secondary and Higher Secondary Marathon

For students of class 11-12 (age 16+)
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Phlembac Adib Hasan
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Secondary and Higher Secondary Marathon

Unread post by Phlembac Adib Hasan » Sat Nov 10, 2012 12:13 pm

Here we'll talk about national level problems.
Rules:
1. You can post any 'math-problem' from anywhere if you are sure it has a solution. Also you should give the source.(Like book name, link or self-made)
2. If a problem remains unsolved for two days, the proposer must post the solution (for self-made problems) or the official solution will be posted. (for contest problems)
3. Anyone can post a new problem iff the previous problem has been solved already.
4. Don't forget to type the problem number.
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SANZEED
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Re: Secondary and Higher Secondary Marathon

Unread post by SANZEED » Sat Nov 10, 2012 2:04 pm

Problem 1
Prove that the expression
$m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$
can't have the value $33$ regardless of what integers are substituted for $m,n$.

Source:The USSR Math Olympiad Problem Book.
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Fahim Shahriar
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Re: Secondary and Higher Secondary Marathon

Unread post by Fahim Shahriar » Sat Nov 10, 2012 5:26 pm

Problem 1 Solution:
Simplifying the expression we get,

$m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$
$=(m+n)(m-n)(m+2n)(m-2n)(m+3n)$

It has five factors. On the other hand, 33 has only two factors. If one factor or the product of two factors of the expression is 33, the other factors will be 1. But then we will get $m=n$ and a factor will be 0.
Therefore, the expression can't have the value 33.
Name: Fahim Shahriar Shakkhor
Notre Dame College

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Fahim Shahriar
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Re: Secondary and Higher Secondary Marathon

Unread post by Fahim Shahriar » Sat Nov 10, 2012 6:16 pm

Problem 2
$ABCDEFGHI$ is a regular nonagon . show that $BG = BC + BD$
Name: Fahim Shahriar Shakkhor
Notre Dame College

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SANZEED
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Re: Secondary and Higher Secondary Marathon

Unread post by SANZEED » Sat Nov 10, 2012 7:57 pm

For Problem $1$
I think I should indicate to a few facts here:
:arrow: $33$ can be expressed as a product of $4$ different factors.
$33=11\cdot (-3)\cdot 1\cdot (-1)$
$33=(-11)\cdot 3\cdot 1\cdot (-1)$.
:arrow: Secondly,my proof is as follows: if $n\neq 0$,then none of the factors are equal. If $n=0$, then the expression becomes $m^5$,which can't be equal to $33$.
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Phlembac Adib Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Phlembac Adib Hasan » Sat Nov 10, 2012 10:03 pm

Hints for prob 2:
WLOG we may assume the nonagon inscribed in the unit circle. Now use complex numbers.
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nafistiham
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Re: Secondary and Higher Secondary Marathon

Unread post by nafistiham » Sun Nov 11, 2012 10:54 am

Phlembac Adib Hasan wrote:Hints for prob 2:
WLOG we may assume the nonagon inscribed in the unit circle. Now use complex numbers.
Yes, unit circles is the trick. But, we may not to have to go to that much complexity. just, simple trigonometry can serve us the solution.

a little trigonometric hint.
$D',C',G'$ are midpoints of $BD,BC,BG$
prove
\[\sin80=\sin20+\sin40\]
salted(nona)gon.png
angles looking like right angles are right angles :)
salted(nona)gon.png (244.64 KiB) Viewed 4807 times
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Fahim Shahriar
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Re: Secondary and Higher Secondary Marathon

Unread post by Fahim Shahriar » Sun Nov 11, 2012 11:32 pm

Problem 2 Solution
nona.jpg
Structure
nona.jpg (14.68 KiB) Viewed 4788 times
:arrow: $BC$ ও $FE$ কে বর্ধিত করি; তারা $M$ বিন্দুতে ছেদ করে। $EC$ and $FB$ are parallel with $HI$.

We can observe that $CE=BD$ and $BF=BG$.

Each angle of the nonagon is $(180-\frac{360}{9}) = 140^o = \angle EDC$

As $ED=CD$, $\angle DEC = \angle DCE = 20^o$
$\angle FEM$ সরলকোণ এবং $\angle FED = 140^o$. So, $\angle DEM = 40^o$. Hence, $\angle MEC = 60^o$. Similarly, $\angle MCE = 60^o$.

$\Delta MEC$ and $\Delta MFB$ both are equilateral. $BF=BM$ and $CE=CM$.
$BG=BF=BM=BC+CM=BC+CE=BC+BD$

$BG=BC+BD$ [Proved] :D


Someone post problem 3.
Name: Fahim Shahriar Shakkhor
Notre Dame College

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Tahmid Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Tahmid Hasan » Mon Nov 12, 2012 12:33 am

Problem 3: Circles $W_1,W_2$ meet at $D$ and $P$. $A$ and $B$ are on $W_1,W_2$ respectively, such that $AB$ is tangent to $W_1$ and $W_2$. Suppose $D$ is closer than $P$ to the line $AB$. $AD$ meet circle $W_2$ for second time at $C$. Let $M$ be the midpoint of $BC$. Prove that $\angle{DPM}=\angle{BDC}$.
Source: Iran NMO-2010-3.
Note: High-school textbook theorems are enough, so I thought it would be okay to post here.
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SANZEED
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Re: Secondary and Higher Secondary Marathon

Unread post by SANZEED » Mon Nov 12, 2012 1:05 am

Problem 3
Since $\angle BPM$ is common,so we just need to show that $\angle DPM=\angle CPM$. Let the extension of $PD$ meet $AB$ at $E$. Now,since $PE$ is the radical axis of the two circles,and $AB$ is tangent to the circles,we must have $EA^2=EB^2\Rightarrow EA=EB$. So $EM\parallel AC$. Thus, $\angle EPB=\angle DPB=\angle DCB=\angle EMB$
and $E,B,M,P$ are con cyclic. Again, $\angle EBD=\angle BCD=\angle BPD$.
Now,in cyclic quad $EBMP$, we have that $\angle EBM+\angle EPM=180^{\circ}$ and in cyclic quad $BDPC$ we have $\angle DBC+\angle DPC=180^{\circ}$.
So, $\angle BPD=\angle EBD=\angle EBM-\angle DBC=180^{\circ}-\angle EPM-180^{\circ}+\angle DPC=\angle CPM$, which completes the proof.

P.S.Someone post problem $4$ please.
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