Secondary and Higher Secondary Marathon

For students of class 11-12 (age 16+)
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SANZEED
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Re: Secondary and Higher Secondary Marathon

Unread post by SANZEED » Sat Jan 05, 2013 1:11 pm

:oops: Forgot to mention the source. Source of my problem : www.brilliant.org
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

sourav das
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Re: Secondary and Higher Secondary Marathon

Unread post by sourav das » Thu Jan 17, 2013 3:01 pm

SANZEED wrote:$\boxed {24}$
Let $N={1,2,3,...,2012}$. A $3$ element subset $S$ of $N$ is called $4$-splittable if there is an $n\in (N-S)$ such that $S\bigcup {n}$ can be partitioned into two sets such that the sum of each set is the same. How many $3$ element subsets of $N$ is non-$4$-splittable?
Solution:
We'll find general solution for $N=\{1,2,3.......n\}$
Let $\{a,b,c\}$ is a non-$4$-splittable subset. W.L.O.G. $n\geq a>b>c \geq 1$ . Now if $n\geq (a+b-c)=k$ then it will be 4 splittable (as, $k=(a+b-c)>a$). So, $a+b-c>n$______(Condition_1)
Note that $a>(b+c-a)=t$ not belongs to $\{a,b,c\}$; otherwise 2 of them will be same. So $t=(b+c-a)<1$ otherwise it will become 4-splittable. So, $(a-b-c)\geq 0$_____________(Condition_2)
Now $a>(a+c-b)>c$ implies $b=a+c-b$. Otherwise, the 4th element will be $(a+c-b)$ to make it 4-splittable.
So, $a+c=2b$_________(Condition_3)
Now, if $a-b-c=0$ then, $b=2c$ and then $a=3c$...........(s_1)
Otherwise $a-b-c=b-2c>0$ and $b>(b-2c)$. So, $b-2c=c$. (otherwise the 4th element will be $b-2c$ and it will become 4-splitable)
So, here $b=3c,a=5c$...........(s_2)
For (s_1) type solutions; note that $c>\frac{n}{4}$ (Condition_1) and $3c=a\leq n$. So, $c\leq \frac{n}{3}$

For (s_2) type solutions; note that $c>\frac{n}{7}$ (Condition_1) and $5c=a\leq n$. So, $c\leq \frac{n}{5}$

So total number of all non-4-splittable sets $\left \lfloor \frac{n}{3} \right \rfloor - \left \lfloor \frac{n}{4} \right \rfloor + \left \lfloor \frac{n}{5} \right \rfloor- \left \lfloor \frac{n}{7} \right \rfloor$

P.S: I have to say "Cool problem"
Anyone can take my turn.
You spin my head right round right round,
When you go down, when you go down down......
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Tahmid Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Tahmid Hasan » Thu Jan 17, 2013 4:51 pm

Problem $25$: Let $\gamma$ be the circumcircle of the acute triangle $ABC$. Let $P$ be the midpoint of the minor arc $BC$. The parallel to $AB$ through $P$ cuts $BC$, $AC$ and $\gamma$ at points $R,S$ and $T$, respectively. Let $K \equiv AP \cap BT$ and $L \equiv BS \cap AR$. Show that $KL$ passes through the midpoint of $AB$ if and only if $CS = PR$.
Source: Centroamerican 2012-2.
বড় ভালবাসি তোমায়,মা

sourav das
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Re: Secondary and Higher Secondary Marathon

Unread post by sourav das » Thu Jan 17, 2013 5:54 pm

Solution:
As, $AB||PT$, $P$ is the midpoint of minor arc $BC$ and using cyclic property : $BP=CP=AT$ and so $AP||CT$ (Note that $ABPT,APCT$ cyclic trapezium and the two non-parallel sides of a cyclic trapezium are equal). Similarly in trapezium $APCT$, $CS=TS$ Now $CS=PR \Leftrightarrow PR=ST \Leftrightarrow \triangle BPR \cong \triangle ATS$
$\Leftrightarrow BR=AS$ in trapezium $BRSA$ $\Leftrightarrow K,L$ are on the perpendicular bisector of $AB$
$\Leftrightarrow KL$ passes through the midpoint of $AB$ (Note that $BP=AT$ in trapezium $ABPT$ and $AP\cap BT=K$)
Anyone can take my turn
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

Reza_raj
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Re: Secondary and Higher Secondary Marathon

Unread post by Reza_raj » Thu Jan 17, 2013 8:39 pm

Can you please explain this!
I don't understand this!

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Tahmid Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Tahmid Hasan » Thu Jan 17, 2013 9:05 pm

Problem $26$: Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.
Source: JBMO-2010-3.
বিঃদ্রঃ বেশি অভিজ্ঞদের(বিশেষ করে সৌরভ ভাই) এই ম্যারাথনে এত তাড়াতাড়ি সমাধান না দেওয়ার অনুরোধ জানাচ্ছি :P
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sourav das
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Re: Secondary and Higher Secondary Marathon

Unread post by sourav das » Thu Jan 17, 2013 11:37 pm

Reza_raj wrote:Can you please explain this!
I don't understand this!
Which part?
@Tahmid, মোটামুটি ৪ মাস পর এই বছরের সমস্যা সমাধান করা শুরু করলাম। নেশার মত লাগতাসে। নেশা...... চেষ্টা করব পরে পোস্ট করার (সমাধান করতে পারলে)
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Phlembac Adib Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Phlembac Adib Hasan » Fri Jan 18, 2013 12:20 pm

Tahmid Hasan wrote:Problem $26$: Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.
Source: JBMO-2010-3.
বিঃদ্রঃ বেশি অভিজ্ঞদের(বিশেষ করে সৌরভ ভাই) এই ম্যারাথনে এত তাড়াতাড়ি সমাধান না দেওয়ার অনুরোধ জানাচ্ছি :P
In $\triangle ABK,$ the angle bisector of $\angle A$ and perpendicular bisector of opposite side, $BK$, meet at point $M$. So $M$ must lie on $\bigcirc ABK$. After some angle chasing, we get $\angle NLA=\angle NAL=\angle B/2$.
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Phlembac Adib Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Phlembac Adib Hasan » Fri Jan 18, 2013 12:31 pm

Problem 27
Find all functions $f:\mathbb R\to \mathbb R$ satisfying this equation:
\[f(xy+f(x))=xf(y)+f(x)\]
Source: An excalibur problem. I solved it in the last month, so can't remember the particular source.

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*Mahi*
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Re: Secondary and Higher Secondary Marathon

Unread post by *Mahi* » Sat Jan 19, 2013 1:36 pm

Phlembac Adib Hasan wrote:Problem 27
Find all functions $f:\mathbb R\to \mathbb R$ satisfying this equation:
\[f(xy+f(x))=xf(y)+f(x)\]
Source: An excalibur problem. I solved it in the last month, so can't remember the particular source.
$f(x) = 0 \forall x \in \mathbb R$ is the only constant solution for the given function.
Let $P(x,y)$ be the assertion \[f(xy+f(x))=xf(y)+f(x)\]
Let $f(x) = f(y) \neq 0$ for some $x,y \in \mathbb R$. Then \[f(xy+f(x))=f(xy+f(y))\] \[\Rightarrow xf(y)+f(x)= yf(x)+f(y)\] \[\Rightarrow x = y\] which implies injectivity.
Now, $P(0,0) \Rightarrow f(f(0)) = f(0)$, or $f(0) = 0$.
And finally $P(x,0) \Rightarrow f(f(x)) = f(x) \Rightarrow f(x) = x \forall x \in \mathbb R$
So, there are two solutions in total $f(x) = 0 \forall x \in \mathbb R$ and $f(x) = x \forall x \in \mathbb R$
Feel free to take my turn in posting problem.
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