Secondary and Higher Secondary Marathon
 Phlembac Adib Hasan
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Secondary and Higher Secondary Marathon
Here we'll talk about national level problems.
Rules:
1. You can post any 'mathproblem' from anywhere if you are sure it has a solution. Also you should give the source.(Like book name, link or selfmade)
2. If a problem remains unsolved for two days, the proposer must post the solution (for selfmade 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.
Rules:
1. You can post any 'mathproblem' from anywhere if you are sure it has a solution. Also you should give the source.(Like book name, link or selfmade)
2. If a problem remains unsolved for two days, the proposer must post the solution (for selfmade 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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Re: Secondary and Higher Secondary Marathon
Problem 1
Prove that the expression
$m^5+3m^4n5m^3n^215m^2n^3+4mn^4+12n^5$
can't have the value $33$ regardless of what integers are substituted for $m,n$.
Source:The USSR Math Olympiad Problem Book.
Prove that the expression
$m^5+3m^4n5m^3n^215m^2n^3+4mn^4+12n^5$
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
 Posts: 138
 Joined: Sun Dec 18, 2011 12:53 pm
Re: Secondary and Higher Secondary Marathon
Problem 1 Solution:
Simplifying the expression we get,
$m^5+3m^4n5m^3n^215m^2n^3+4mn^4+12n^5$
$=(m+n)(mn)(m+2n)(m2n)(m+3n)$
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.
Simplifying the expression we get,
$m^5+3m^4n5m^3n^215m^2n^3+4mn^4+12n^5$
$=(m+n)(mn)(m+2n)(m2n)(m+3n)$
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
Notre Dame College
 Fahim Shahriar
 Posts: 138
 Joined: Sun Dec 18, 2011 12:53 pm
Re: Secondary and Higher Secondary Marathon
Problem 2
$ABCDEFGHI$ is a regular nonagon . show that $BG = BC + BD$
$ABCDEFGHI$ is a regular nonagon . show that $BG = BC + BD$
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
Re: Secondary and Higher Secondary Marathon
For Problem $1$
I think I should indicate to a few facts here:
$33$ can be expressed as a product of $4$ different factors.
$33=11\cdot (3)\cdot 1\cdot (1)$
$33=(11)\cdot 3\cdot 1\cdot (1)$.
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 $33$.
I think I should indicate to a few facts here:
$33$ can be expressed as a product of $4$ different factors.
$33=11\cdot (3)\cdot 1\cdot (1)$
$33=(11)\cdot 3\cdot 1\cdot (1)$.
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
Hints for prob 2:
WLOG we may assume the nonagon inscribed in the unit circle. Now use complex numbers.
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
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.Phlembac Adib Hasan wrote:Hints for prob 2:
WLOG we may assume the nonagon inscribed in the unit circle. Now use complex numbers.
a little trigonometric hint.
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please cooperate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please cooperate.
 Fahim Shahriar
 Posts: 138
 Joined: Sun Dec 18, 2011 12:53 pm
Re: Secondary and Higher Secondary Marathon
Problem 2 Solution
Someone post problem 3.
$BC$ ও $FE$ কে বর্ধিত করি; তারা $M$ বিন্দুতে ছেদ করে। $EC$ and $FB$ are parallel with $HI$.
Someone post problem 3.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
 Tahmid Hasan
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 Location: Khulna,Bangladesh.
Re: Secondary and Higher Secondary Marathon
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 NMO20103.
Note: Highschool textbook theorems are enough, so I thought it would be okay to post here.
Source: Iran NMO20103.
Note: Highschool textbook theorems are enough, so I thought it would be okay to post here.
বড় ভালবাসি তোমায়,মা
Re: Secondary and Higher Secondary Marathon
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, $\angle BPD=\angle EBD=\angle EBM\angle DBC=180^{\circ}\angle EPM180^{\circ}+\angle DPC=\angle CPM$, which completes the proof.
P.S.Someone post problem $4$ please.
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, $\angle BPD=\angle EBD=\angle EBM\angle DBC=180^{\circ}\angle EPM180^{\circ}+\angle DPC=\angle CPM$, which completes the proof.
P.S.Someone post problem $4$ please.
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