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.

Re: Secondary and Higher Secondary Marathon

Posted: Sat Nov 10, 2012 2:04 pm

by SANZEED

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.

Re: Secondary and Higher Secondary Marathon

Posted: Sat Nov 10, 2012 5:26 pm

by Fahim Shahriar

Problem 1 Solution:
Simplifying the expression we get,

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.

Re: Secondary and Higher Secondary Marathon

Posted: Sat Nov 10, 2012 6:16 pm

by Fahim Shahriar

Problem 2
$ABCDEFGHI$ is a regular nonagon . show that $BG = BC + BD$

Re: Secondary and Higher Secondary Marathon

Posted: Sat Nov 10, 2012 7:57 pm

by SANZEED

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 $33$.

Re: Secondary and Higher Secondary Marathon

Posted: Sat Nov 10, 2012 10:03 pm

by Phlembac Adib Hasan

Hints for prob 2:
WLOG we may assume the nonagon inscribed in the unit circle. Now use complex numbers.

Re: Secondary and Higher Secondary Marathon

Posted: Sun Nov 11, 2012 10:54 am

by nafistiham

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
\[\sin80=\sin20+\sin40\]

angles looking like right angles are right angles :)

salted(nona)gon.png (244.64 KiB) Viewed 4809 times

Re: Secondary and Higher Secondary Marathon

Posted: Sun Nov 11, 2012 11:32 pm

by Fahim Shahriar

Problem 2 Solution

Structure

nona.jpg (14.68 KiB) Viewed 4790 times

$BC$ 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]

Someone post problem 3.

Re: Secondary and Higher Secondary Marathon

Posted: Mon Nov 12, 2012 12:33 am

by Tahmid Hasan

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.

Re: Secondary and Higher Secondary Marathon

Posted: Mon Nov 12, 2012 1:05 am

by SANZEED

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.