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 \(x+y+z\).

Check out all my notes and stuffs for more math problems!

## Comments

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TopNewestIs the first answer -1?? – Abhineet Nayyar · 2 years ago

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– Samuraiwarm Tsunayoshi · 2 years ago

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

5th - answer 3 – Krishna Sharma · 2 years ago

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– Samuraiwarm Tsunayoshi · 2 years ago

Yep, but how will you prove that no.5 has exactly 1 answer?Log in to reply

– Aneesh Kundu · 1 year, 12 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.Log in to reply