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)(1x+1y+1z)=1(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(x+y)(y+z)(z+x) = 0.

3.) Find all roots of x2x3=x7\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)P(x) with real coefficients such that P(1)=210P(1) = 210 and

(x+10)P(2x)=(8x32)P(x+6)(x+10)P(2x) = (8x-32)P(x+6).

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

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Thailand Math POSN 2013

Thailand Math POSN 2014

Note by Samuraiwarm Tsunayoshi
6 years ago

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Is the first answer -1??

Abhineet Nayyar - 6 years ago

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Nope there's another case of a,b,c,d you didn't consider.

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For1st I got 3

5th - answer 3

Krishna Sharma - 6 years ago

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@Krishna Sharma Yep, but how will you prove that no.5 has exactly 1 answer?

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@Samuraiwarm Tsunayoshi We can apply AM-GM first on x,y,zx,y,z and then on xy,yz,zx\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.

Aneesh Kundu - 5 years, 12 months ago

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