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Thread: Mongolia's province math competition 9th grade.

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    Question 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
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    Re: Mongolia's province math competition 9th grade.

    Quote Originally Posted by anod1996 View Post
    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
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    Re: Mongolia's province math competition 9th grade.

    Quote Originally Posted by anod1996 View Post
    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:
    $\displaystyle 2^(2k+1)+9*2^k+5 = 2*(2^k)^2+9*2^k+5 = 2a^2+9a+5$ [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.
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