G3.1 In $\triangle A B C, \angle A B C=2 \angle A C B, B C=2 A B$. If $\angle B A C=a^{\circ}$, find the value of $a$.

G3.2 Given that $x+\frac{1}{x}=\sqrt{2}, \frac{x^{2}}{x^{4}+x^{2}+1}=b$, find the value of $b$.
G3.3 If the number of positive integral root(s) of the equation $x+y+2 x y=141$ is $c$, find the value of $c$.

G3.4 Given that $x+y+z=0, x^{2}+y^{2}+z^{2}=1$ and $d=2\left(x^{4}+y^{4}+z^{4}\right)$, find the value of $d$.
Let $\angle A C B=\theta, \angle A B C=2 \theta$ (given)
$A B=c, B C=2 c$
$\angle B A C=180^{\circ}-\theta-2 \theta(\angle \mathrm{s}$ sum of $\triangle)$
By sine formula, $\frac{c}{\sin \theta}=\frac{2 c}{\sin \left(180^{\circ}-3 \theta\right)}$
$$
\begin{array}{l}
\sin 3 \theta=2 \sin \theta \\
3 \sin \theta-4 \sin ^{3} \theta=2 \sin \theta \\
4 \sin ^{2} \theta-1=0 \\
\sin \theta=\frac{1}{2} ; \theta=30^{\circ}, \angle B A C=180^{\circ}-3 \theta=90^{\circ} ; a=90
\end{array}
$$

Method 2 Let $\angle A C B=\theta, \angle A B C=2 \theta$ (given)
Let $S$ be the mid-point of $B C$.
Let $N$ and $M$ be the feet of perpendiculars drawn from $S$ on $A C$ and $B$ from $A S$ respectively.
$\triangle B S M \cong \triangle B A M$ (RHS)
$\angle R Q N=\theta=\angle S Q N$ (corr. $\angle \mathrm{s}, \cong \Delta$ 's)
$\triangle C S N \cong \triangle B S M \cong \triangle B A M$ (AAS)
$N S=M S=A M$ (corr. sides $\cong \Delta ' s$ s)
$\sin \angle N A S=\frac{N S}{A S}=\frac{1}{2} ; \angle N A S=30^{\circ}$;
$\angle A S N=60^{\circ}(\angle \mathrm{s}$ sum of $\triangle A S N)$
$90^{\circ}-\theta+60^{\circ}+90^{\circ}-\theta=180^{\circ}(\mathrm{adj} . \angle \mathrm{s}$ on st. line $B S C$ )
$\theta=30^{\circ}$
$\angle B A C=180^{\circ}-3 \theta=90^{\circ}(\angle \mathrm{s}$ sum of $\triangle A B C)$
$a=90$
$\left(x+\frac{1}{x}\right)^{2}=2 \Rightarrow x^{2}+2+\frac{1}{x^{2}}=2$
$\Rightarrow x^{2}+\frac{1}{x^{2}}=0$ (remark: $x$ is a complex number)
$b=\frac{x^{2}}{x^{4}+x^{2}+1}=\frac{1}{x^{2}+1+\frac{1}{x^{2}}}=1$
$2 x+2 y+4 x y=282 \Rightarrow 2 x+2 y+4 x y+1=283$, which is a prime number
$(2 x+1)(2 y+1)=1 \times 283$
$2 x+1=1,2 y+1=283$ (or $2 x+1=283,2 y+1=1$ )
Solving the above equations, there is no positive integral roots.
$$
c=0
$$

Let $x+y+z=0 \ldots \ldots$ (1), $x^{2}+y^{2}+z^{2}=1 \ldots \ldots$ (2)
From (1), $(x+y+z)^{2}=0$
$$
\Rightarrow x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=0
$$

Sub. (2) into the above equation, $x y+y z+z x=-\frac{1}{2}$
From (3), $(x y+y z+z x)^{2}=\frac{1}{4}$
$$
\Rightarrow x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+2 x y z(x+y+z)=\frac{1}{4}
$$

Sub. (1) into the above equation, $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}=\frac{1}{4}$
From (2), $\left(x^{2}+y^{2}+z^{2}\right)^{2}=1$
$$
\Rightarrow x^{4}+y^{4}+z^{4}+2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)=1
$$

Sub. (4) into the above equation, $x^{4}+y^{4}+z^{4}=\frac{1}{2}$
Sub. (5) into $d \Rightarrow d=2\left(x^{4}+y^{4}+z^{4}\right)=2 \times \frac{1}{2}=1$