Secondary and Higher Secondary Marathon

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

Unread post by Tahmid Hasan » Sat Jan 19, 2013 2:06 pm

Problem $28$: Let $ABCD$ be a cyclic quadrilateral with opposite sides not parallel. Let $X$ and $Y$ be the
intersections of $AB,CD$ and $AD,BC$ respectively. Let the angle bisector of $\angle AXD$ intersect $AD,BC$ at $E,F$ respectively, and let the angle bisectors of $\angle AYB$ intersect $AB,CD$ at $G,H$ respectively. Prove that $EFGH$ is a parallelogram.
Source:Canada National-2011-2
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Nadim Ul Abrar
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Re: Secondary and Higher Secondary Marathon

Unread post by Nadim Ul Abrar » Sat Jan 19, 2013 8:32 pm

28 .
$\displaystyle YA.YD=YB.YC \rightarrow \frac {YA}{YB}=\frac{YC}{YD} \rightarrow \frac {AG}{BG}=\frac{CH}{DH}$
Similarly , $\displaystyle \frac {AE}{DE}=\frac{CF}{BF}$

Using sine rule $\displaystyle \frac {YA}{YB}=\frac{XA}{XB}=\frac {sin \angle YAB}{sin \angle YBA} \rightarrow
\frac {AG}{BG}=\frac {AE}{DE} \rightarrow \frac{CH}{DH}=\frac{CF}{BF} \rightarrow GE||BD||HF $

And $\displaystyle \frac {BG}{AG}=\frac{DE}{AE}=\frac{BF}{CF}=\frac{DH}{CH} \rightarrow GF||AC||EH $
$\frac{1}{0}$

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Nadim Ul Abrar
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Re: Secondary and Higher Secondary Marathon

Unread post by Nadim Ul Abrar » Sat Jan 19, 2013 9:01 pm

Fun: For problem 28 , EFGH is রম্বস as well .

Problem 29 : Find all positive integers $n$ and $k$ such that $(n+1)^n=2n^k+3n+1$
(spain '12 D2 1)
$\frac{1}{0}$

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SANZEED
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Re: Secondary and Higher Secondary Marathon

Unread post by SANZEED » Sat Jan 19, 2013 11:24 pm

$\boxed {29}$
We have $(n+1)^{n}-1=2n^{k}+3n\Rightarrow n^{n}+\binom{n}{1}n^{n-1}+...+\binom{n}{n-1}n+1-1=2n^{k}+3n$
Since $n$ is a positive integer,we can divide both sides by $n$. This implies,
$n^{n-1}+\binom{n}{1}n^{n-2}+...+\binom{n}{n-1}=2n^{k-1}+3$. Now in the last equation $n$ divides the left hand side,and also it divides the first term of the right hand side. So we must have $n|3$. Thus the possible values for $n$ are $1,3$. Checking implies that $(n,k)=(3,3)$ is the only solution.
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Tahmid Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Tahmid Hasan » Sun Jan 20, 2013 12:07 pm

Since Sanzeed didn't post a new problem. I'm posting one.
Problem $30$: Let $P$ be a point inside a square $ABCD$ such that $PA=1,PB=2,PC=3$. The area of $ABCD$ can be expressed as $a+b\sqrt c$ where $a,b,c \in \mathbb{N}$ and $c$ is not divisible by the square of any prime. What is the value of $a+b+c$?
Source: Baltic Way-2011-12(A little edited by me 8-) )
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zadid xcalibured
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Re: Secondary and Higher Secondary Marathon

Unread post by zadid xcalibured » Sun Jan 20, 2013 2:13 pm

$a=5$,$b=-2$,$c=2$ So $a+b+c=5$
I have a shitty proof.

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Tahmid Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Tahmid Hasan » Sun Jan 20, 2013 3:45 pm

Here's my solution- A rotation of $90^{\circ}$ in clockwise direction sends $\triangle BAP$ to $\triangle BCQ$, $P,Q$ are correspondent points.
So $\angle PBQ=90^{\circ}$ and $BP=BQ=2,AP=CQ=1$. Hence $\angle PQB=45^{\circ}$.
By Pythagoras' theorem on $\triangle BPQ,PQ=2\sqrt 2$.
By the converse of Pythagoras on $\triangle CPQ, \angle PQC=90^{\circ}$
So $\angle BQC=135^{\circ}$.
Hence $BC^2=BQ^2+CQ^2-2.BQ.CQ.\cos BQC=5+2\sqrt 2$.
So $a+b+c=9$.
Really cool problem 8-)
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Last edited by Tahmid Hasan on Sun Jan 20, 2013 7:47 pm, edited 1 time in total.
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zadid xcalibured
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Re: Secondary and Higher Secondary Marathon

Unread post by zadid xcalibured » Sun Jan 20, 2013 5:15 pm

Oy Tahmid,This is my shitty solution.I find it so uncool. ;)
The mistake in my solution is that I took the positive value of some cos when it should be negative. :x

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FahimFerdous
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Re: Secondary and Higher Secondary Marathon

Unread post by FahimFerdous » Sun Jan 20, 2013 6:18 pm

In 2012 national camp, Sourav solved a slightly different version of this problem using some cool rotation! Mugdho vaia solved it using co-ordinates though.
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Tahmid Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Tahmid Hasan » Sun Jan 20, 2013 7:30 pm

zadid xcalibured wrote:Oy Tahmid,This is my shitty solution.I find it so uncool. ;)
Then surely we have different definitions of cool :|
FahimFerdous wrote:In 2012 national camp, Sourav solved a slightly different version of this problem using some cool rotation!
I have heard from Labib vai that his(Sourav vai's) solution was one of the coolest things shown in 2012 national camp, but I never had the opportunity saw it :(
Actually I totally forgot about about this problem, I was solving some 'Mathematical Excalibur' problems the other day when I came by this problem
Inside an equilateral triangle $ABC$, there is a point $P$ such that $PC=3,PA=4$ and $PB=5$. Find the
perimeter of $\triangle ABC$.
The solution was simply amazing and gave me the incentives to solve this one :)
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