Page **9** of **13**

### Re: Secondary and Higher Secondary Marathon

Posted: **Sat Jan 19, 2013 2:06 pm**

by **Tahmid Hasan**

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

### Re: Secondary and Higher Secondary Marathon

Posted: **Sat Jan 19, 2013 8:32 pm**

by **Nadim Ul Abrar**

### Re: Secondary and Higher Secondary Marathon

Posted: **Sat Jan 19, 2013 9:01 pm**

by **Nadim Ul Abrar**

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)

### Re: Secondary and Higher Secondary Marathon

Posted: **Sat Jan 19, 2013 11:24 pm**

by **SANZEED**

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 12:07 pm**

by **Tahmid Hasan**

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

)

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 2:13 pm**

by **zadid xcalibured**

$a=5$,$b=-2$,$c=2$ So $a+b+c=5$

I have a shitty proof.

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 3:45 pm**

by **Tahmid Hasan**

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

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 5:15 pm**

by **zadid xcalibured**

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.

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 6:18 pm**

by **FahimFerdous**

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.

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 7:30 pm**

by **Tahmid Hasan**

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