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# Algebra (Thailand Math POSN 1st elimination round 2014)

Way too easier than last year.

Write the full solution.

1.) Let $$a,b,c,d$$ be real numbers such that $$\displaystyle \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{a}$$. Find the maximum value of $$\displaystyle \frac{ab-3bc+ca}{a^{2}-b^{2}+c^{2}}$$.

2.) If $$(x+y+z)\left(\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) = 1$$, prove that $$(x+y)(y+z)(z+x) = 0$$.

3.) Find all roots of $$\displaystyle \left\lfloor \frac{x}{2} \right\rfloor - \left\lfloor \frac{x}{3} \right\rfloor = \frac{x}{7}$$.

4.) (Someone posted this problem before in Brilliant.) Find all polynomials $$P(x)$$ with real coefficients such that $$P(1) = 210$$ and

$(x+10)P(2x) = (8x-32)P(x+6)$.

5.) If $$x,y,z$$ are positive real numbers such that $$\displaystyle x+\frac{y}{z} = y+\frac{z}{x} = z+\frac{x}{y} = 2$$. Find the value of $$x+y+z$$.

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Thailand Math POSN 2013

Thailand Math POSN 2014

Note by Samuraiwarm Tsunayoshi
2 years, 2 months ago

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Is the first answer -1?? · 2 years, 2 months ago

Nope there's another case of a,b,c,d you didn't consider. · 2 years, 2 months ago

For1st I got 3

5th - answer 3 · 2 years, 2 months ago

Yep, but how will you prove that no.5 has exactly 1 answer? · 2 years, 2 months ago

We can apply AM-GM first on $$x,y,z$$ and then on $$\frac{x}{y},\frac{y}{z},\frac{z}{x}$$ and add both and say that since it is at its minimum so there is equality between the terms. · 2 years, 2 months ago