1. 3^{x}+4^{y}=5^{z}find all natural x, y, z numbers

2. 2^{2k+1}+9*2^{k}+5=n^{2}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

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- Jan 25th 2013, 02:59 AManod1996Mongolia's province math competition 9th grade.
1. 3

^{x}+4^{y}=5^{z}find all natural x, y, z numbers

2. 2^{2k+1}+9*2^{k}+5=n^{2}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, 07:11 AMtopsquarkRe: Mongolia's province math competition 9th grade.
- Jan 25th 2013, 10:42 AMearthboyRe: Mongolia's province math competition 9th grade.
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:

$\displaystyle 2^(2k+1)+9*2^k+5 = 2*(2^k)^2+9*2^k+5 = 2a^2+9a+5$ [taking $\displaystyle a = 2^k$]

=(#now we complete the squares) $\displaystyle 4a^2+2*2a*2+4-(2a^2-a-1)$

$\displaystyle =(2a+2)^2-(2a^2-a-1)$

for the expression to be a perfect square, $\displaystyle 2a^2-a-1=0$, which happens when $\displaystyle a=-1/2 , 1$.as k is integer, we take 1

so $\displaystyle 2^k=1$, hence k can only be $\displaystyle 0$

3.This problem is both tricky and famous(& i'm giving the most well known solution):

consider $\displaystyle \angle PCB = x$ and we know $\displaystyle \angle PBC + \angle PCB =80$, hence$\displaystyle \angle PBC= 80-x$

now we have to use the trigonometric version of ceva's theorem:

$\displaystyle \frac{\sin \angle PBA}{\sin \angle PAB}*\frac{\sin \angle PCB}{\sin \angle PBC}*\frac{\sin \angle PAC}{\sin \angle PCA} =1$

=> $\displaystyle \frac{\sin 20 \sin x \sin 40}{\sin 10 \sin (80-x) \sin 30}=1$

=>$\displaystyle \frac{4\sin x \sin 40 \cos 10}{\sin (80-x)}=1$

=>$\displaystyle \frac{2\sin x(\sin 30+\sin 50)}{\sin (80-x)}=1$ [#using product-sum formula]

=>$\displaystyle \frac{\sin x(1+2\cos 40)}{\sin (80-x)}=1$

=>$\displaystyle 2\sin x \cos 40 = \sin (80-x)-\sin x = 2\sin(40-x)\cos 40$

so as $\displaystyle 2\sin x \cos 40 = 2\sin(40-x)\cos 40$, we can write $\displaystyle x=40-x$ or $\displaystyle x=20$

hence $\displaystyle \angle ACB=\angle ABC = 50$,therefore triangle ABC is isosceles.