Secondary and Higher Secondary Marathon
 Tahmid Hasan
 Posts: 665
 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
Re: Secondary and Higher Secondary Marathon
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 National20112
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 National20112
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 Nadim Ul Abrar
 Posts: 244
 Joined: Sat May 07, 2011 12:36 pm
 Location: B.A.R.D , kotbari , Comilla
 Nadim Ul Abrar
 Posts: 244
 Joined: Sat May 07, 2011 12:36 pm
 Location: B.A.R.D , kotbari , Comilla
Re: Secondary and Higher Secondary Marathon
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)
Problem 29 : Find all positive integers $n$ and $k$ such that $(n+1)^n=2n^k+3n+1$
(spain '12 D2 1)
$\frac{1}{0}$
Re: Secondary and Higher Secondary Marathon
$\boxed {29}$
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with}} \color{green}{\textit{AVADA KEDAVRA!}}$
 Tahmid Hasan
 Posts: 665
 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
Re: Secondary and Higher Secondary Marathon
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 Way201112(A little edited by me )
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 Way201112(A little edited by me )
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 zadid xcalibured
 Posts: 217
 Joined: Thu Oct 27, 2011 11:04 am
 Location: mymensingh
Re: Secondary and Higher Secondary Marathon
$a=5$,$b=2$,$c=2$ So $a+b+c=5$
I have a shitty proof.
I have a shitty proof.
 Tahmid Hasan
 Posts: 665
 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
Re: Secondary and Higher Secondary Marathon
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 $BC^2=BQ^2+CQ^22.BQ.CQ.\cos BQC=5+2\sqrt 2$.
So $a+b+c=9$.
Really cool problem
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 $BC^2=BQ^2+CQ^22.BQ.CQ.\cos BQC=5+2\sqrt 2$.
So $a+b+c=9$.
Really cool problem
 Attachments

 Sorry, forgot to attach the image!
 BW201112.png (6.92 KiB) Viewed 4221 times
Last edited by Tahmid Hasan on Sun Jan 20, 2013 7:47 pm, edited 1 time in total.
বড় ভালবাসি তোমায়,মা
 zadid xcalibured
 Posts: 217
 Joined: Thu Oct 27, 2011 11:04 am
 Location: mymensingh
Re: Secondary and Higher Secondary Marathon
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.
The mistake in my solution is that I took the positive value of some cos when it should be negative.
 FahimFerdous
 Posts: 176
 Joined: Thu Dec 09, 2010 12:50 am
 Location: Mymensingh, Bangladesh
Re: Secondary and Higher Secondary Marathon
In 2012 national camp, Sourav solved a slightly different version of this problem using some cool rotation! Mugdho vaia solved it using coordinates though.
Your hot head might dominate your good heart!
 Tahmid Hasan
 Posts: 665
 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
Re: Secondary and Higher Secondary Marathon
Then surely we have different definitions of coolzadid xcalibured wrote:Oy Tahmid,This is my shitty solution.I find it so uncool.
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 itFahimFerdous wrote:In 2012 national camp, Sourav solved a slightly different version of this problem using some cool rotation!
Actually I totally forgot about about this problem, I was solving some 'Mathematical Excalibur' problems the other day when I came by this problem
The solution was simply amazing and gave me the incentives to solve this oneInside an equilateral triangle $ABC$, there is a point $P$ such that $PC=3,PA=4$ and $PB=5$. Find the
perimeter of $\triangle ABC$.
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