G3.1 If $\frac{2002^{3}+4 \times 2002^{2}+6006}{2002^{2}+2002}=a$, find the value of $a$.
G3.2 It is given that the real numbers $x$ and $y$ satisfy the relation $y=\frac{x}{2 x-1}$.
G3.3 Suppose two different numbers are chosen randomly from the 50 positive integers $1,2,3, \ldots$, 50 , and the sum of these two numbers is not less than 50 . If the number of ways of choosing these two numbers is $c$, find the value of $c$.

G3.4 Given that $x-y=1+\sqrt{5}, y-z=1-\sqrt{5}$. If $x^{2}+y^{2}+z^{2}-x y-y z-z x=d$, find the value of $d$.
$$
\begin{aligned}
a & =\frac{2002\left(2002^{2}+4 \times 2002+3\right)}{2002(2002+1)} \\
& =\frac{(2002+1)(2002+3)}{2002+1} \\
& =2005
\end{aligned}
$$

If the minimum value of $\frac{1}{x^{2}}+\frac{1}{y^{2}}$ is $b$, find the value of $b$.
$$
\begin{aligned}
\frac{1}{x^{2}}+\frac{1}{y^{2}} & =\frac{1}{x^{2}}+\frac{(2 x-1)^{2}}{x^{2}} \\
& =\frac{4 x^{2}-4 x+2}{x^{2}}
\end{aligned}
$$

Let $T=\frac{4 x^{2}-4 x+2}{x^{2}}$
$$
\begin{array}{l}
T x^{2}=4 x^{2}-4 x+2 \\
(T-4) x^{2}+4 x-2=0 \\
\Delta=4^{2}+4 \times 2(T-4) \geq 0 \\
2+T-4 \geq 0 \\
\Rightarrow T \geq 2
\end{array}
$$

The minimum value is 2
$$
b=2
$$

Reference: 2011 FG2.2
Possible combinations may be: $(1,49),(1,50)$,
$$
(2,48),(2,49),(2,50) \text {, }
$$
$$
\begin{array}{l}
(24,26),(24,27), \ldots,(24,50), \\
(25,26),(25,27), \ldots,(25,50), \\
(26,27), \ldots,(26,50) \\
\begin{aligned}
(49,50)
\end{aligned} \\
\begin{array}{l} 
\text { Total number of combinations }=(2+3+\ldots+25)+25+(24+23+\ldots+1) \\
=(1+2+\ldots+24) \times 2+24+25 \\
2 d=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=(1+\sqrt{5})^{2}+(1-\sqrt{5})^{2}+[(z-y)-(x-y)]^{2} \\
2 d=1+2 \sqrt{5}+5+1-2 \sqrt{5}+5+[-1+\sqrt{5}-(1+\sqrt{5})]^{2}=12+4=16 \\
d=8
\end{array}
\end{array}
$$