Mongolia's province math competition 9th grade.

• Jan 25th 2013, 03:59 AM
anod1996
Mongolia's province math competition 9th grade.
1. 3x+4y=5z find all natural x, y, z numbers
2. 22k+1+9*2k+5=n2 k=? k is an integer
3. If there's a point P inside the triangle ABC where PAB=10 PBA=20 PCA=30 PAC=40 prove that triangle ABC is isosceles
• Jan 25th 2013, 08:11 AM
topsquark
Re: Mongolia's province math competition 9th grade.
Quote:

Originally Posted by anod1996
1. 3x+4y=5z find all natural x, y, z numbers
2. 22k+1+9*2k+5=n2 k=? k is an integer
3. If there's a point P inside the triangle ABC where PAB=10 PBA=20 PCA=30 PAC=40 prove that triangle ABC is isosceles

Surely it can't be legal for you to ask for help on this.

-Dan
• Jan 25th 2013, 11:42 AM
earthboy
Re: Mongolia's province math competition 9th grade.
Quote:

Originally Posted by anod1996
1. 3x+4y=5z find all natural x, y, z numbers
2. 22k+1+9*2k+5=n2 k=? k is an integer
3. If there's a point P inside the triangle ABC where PAB=10 PBA=20 PCA=30 PAC=40 prove that triangle ABC is isosceles

I guess the competition is over or the problems are from old exams....
btw, the problems listed are very common olympiad problems.....
1.Common problem, with a little long solution, got posed on Nick's mathematical puzzle number:98. I am posting the link:
Nick's Mathematical Puzzles: Solution 98

2. we have to make the given expression a square number:
$2^(2k+1)+9*2^k+5 = 2*(2^k)^2+9*2^k+5 = 2a^2+9a+5$ [taking $a = 2^k$]
=(#now we complete the squares) $4a^2+2*2a*2+4-(2a^2-a-1)$
$=(2a+2)^2-(2a^2-a-1)$
for the expression to be a perfect square, $2a^2-a-1=0$, which happens when $a=-1/2 , 1$.as k is integer, we take 1
so $2^k=1$, hence k can only be $0$

3.This problem is both tricky and famous(& i'm giving the most well known solution):
consider $\angle PCB = x$ and we know $\angle PBC + \angle PCB =80$, hence $\angle PBC= 80-x$
now we have to use the trigonometric version of ceva's theorem:
$\frac{\sin \angle PBA}{\sin \angle PAB}*\frac{\sin \angle PCB}{\sin \angle PBC}*\frac{\sin \angle PAC}{\sin \angle PCA} =1$
=> $\frac{\sin 20 \sin x \sin 40}{\sin 10 \sin (80-x) \sin 30}=1$
=> $\frac{4\sin x \sin 40 \cos 10}{\sin (80-x)}=1$
=> $\frac{2\sin x(\sin 30+\sin 50)}{\sin (80-x)}=1$ [#using product-sum formula]
=> $\frac{\sin x(1+2\cos 40)}{\sin (80-x)}=1$
=> $2\sin x \cos 40 = \sin (80-x)-\sin x = 2\sin(40-x)\cos 40$
so as $2\sin x \cos 40 = 2\sin(40-x)\cos 40$, we can write $x=40-x$ or $x=20$
hence $\angle ACB=\angle ABC = 50$,therefore triangle ABC is isosceles.