
Problem 8.1. Let three numbers $a, b$ and $c$ be chosen so that $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}$.
a.) Prove that $a=b=c$.

b.) Find the $\operatorname{sum} x+y$ if $\frac{x}{3 y}=\frac{y}{2 x-5 y}=\frac{6 x-15 y}{x}$ and the expression $-4 x^{2}+36 y-8$ has its maximum value.


## Solution:

a.) It is obvious that $a \neq 0, b \neq 0, c \neq 0$. The first equality gives $b^{2}=a c$, whence by multiplying both sides by $b$ we get $b^{3}=a b c$. Similarly $a^{3}=a b c$ and $c^{3}=a b c$. Hence $a^{3}=b^{3}=c^{3}$ and therefore $a=b=c$.

b.) By multiplying both the numerator and the denominator of the second fraction by 3 and using the result of a.) we obtain $x=3 y$. Thus $-4 x^{2}+36 y-8=-9\left(4 y^{2}-4 y+1\right)+1=$ $-9(2 y-1)^{2}+1$, and its maximum value is 1 when $2 y-1=0$. Therefore $y=\frac{1}{2}$ and $x=\frac{3}{2}$, i.e., $x+y=2$.
