I3.1 A rectangular piece of paper is folded into the following figure. If the area of $\triangle A B C$ is $\frac{1}{3}$ of the area of the original rectangular piece of paper, find the value of $P$.

I3.2 If $Q$ is the positive integral solution of the equation $\frac{P}{2}\left(4^{x}+4^{-x}\right)-35\left(2^{x}+2^{-x}\right)+62=0$, find the value of $Q$.

I3.3 Let $[a]$ be the largest integer not greater than a. For example, $[2.5]=2$. If $R=[\sqrt{1}]+[\sqrt{2}]+\ldots+[\sqrt{99 Q}]$, find the value of $R$.

I3.4 In a convex polygon, other than the interior angle $A$, the sum of all the remaining interior angles is equal to $4 R^{\circ}$. If $\angle A=S^{\circ}$, find the value of $S$.
$$
\begin{array}{l}
B C=P-4, A C=10, \angle A C B=90^{\circ} \\
\frac{(P-4) \cdot 10}{2}=\frac{1}{3} \times P \times 10 \\
\Rightarrow P=12
\end{array}
$$

Let $t=2^{x}+2^{-x}$, then $t^{2}=4^{x}+4^{-x}+2$
$$
\Rightarrow 4^{x}+4^{-x}=t^{2}-2
$$

The equation becomes $6\left(t^{2}-2\right)-35 t+62=0$
$$
\begin{array}{l}
6 t^{2}-35 t+50=0 \\
(2 t-5)(3 t-10)=0 \\
t=\frac{5}{2} \text { or } \frac{10}{3} \\
2^{x}+2^{-x}=\frac{5}{2} \text { or } 2^{x}+2^{-x}=\frac{10}{3} \\
2^{x}+\frac{1}{2^{x}}=\frac{5}{2} \text { or } 2^{x}+\frac{1}{2^{x}}=\frac{10}{3} \\
2\left(2^{x}\right)^{2}+2=5\left(2^{x}\right) \text { or } 3\left(2^{x}\right)^{2}+3=10\left(2^{x}\right) \\
2\left(2^{x}\right)^{2}-5\left(2^{x}\right)+2=0 \text { or } 3\left(2^{x}\right)^{2}-10\left(2^{x}\right)+3=0 \\
\left(2 \cdot 2^{x}-1\right)\left(2^{x}-2\right)=0 \text { or }\left(3 \cdot 2^{x}-1\right)\left(2^{x}-3\right)=0
\end{array}
$$
$2^{x}=\frac{1}{2}, 2, \frac{1}{3}$ or 3
For positive integral solution $x=1 ; Q=1$
$$
\begin{array}{l}
R=[\sqrt{1}]+[\sqrt{2}]+\ldots+[\sqrt{99}]=1+1+1+\underbrace{2+\cdots+2}_{5 \text { times }}+\underbrace{3+\cdots+3}_{7 \text { times }}+\cdots+\underbrace{9+\cdots+9}_{19 \text { times }} \\
R=3 \times 1+5 \times 2+7 \times 3+\ldots+19 \times 9 \\
R=(2 \times 1+1) \times 1+(2 \times 2+1) \times 2+(2 \times 3+1) \times 3+\ldots+(2 \times 9+1) \times 9 \\
R=2 \times 1^{2}+1+2 \times 2^{2}+2+2 \times 3^{2}+3+\ldots+2 \times 9^{2}+9 \\
R=2 \times\left(1^{2}+2^{2}+3^{2}+\ldots+9^{2}\right)+(1+2+3+\ldots+9) \\
R=2 \times \frac{1}{6} \cdot 9(9+1)(2 \times 9+1)+\frac{(1+9) 9}{2}=3 \times 10 \times 19+45=570+45=615
\end{array}
$$

Reference: 1989 HG2, 1990 FG10.3-4, 1992 HG3, 2013 HI6
$$
\begin{array}{l}
4 \times 615+S=180 \times(n-2) \\
S=180(n-2)-2460
\end{array}
$$
$\because$ The polygon is convex
$$
\therefore S<180 . S=180(14)-2460=60
$$