\noindent\textbf{6} \hfill \textit{(See the solution/explanation page)}

\medskip

\noindent In the coordinate space with origin O, consider the cube defined by the inequalities $|x| \leq 1$, $|y| \leq 1$, $|z| \leq 1$. Let $S$ be the part of the surface of this cube satisfying $z < 1$.

\medskip

\noindent Hereafter, when two points A and B in the coordinate space coincide, the line segment AB is defined to represent the point A, and its length is defined to be $0$.

\medskip

\begin{enumerate}
\item[(1)] When a point P in the coordinate space satisfies both of the following conditions (i) and (ii), find the volume $V$ of the region that P can occupy.

\begin{itemize}
\item[(i)] $\mathrm{OP} \leq \sqrt{3}$
\item[(ii)] The line segment OP and $S$ have no common point, or have only the point P as a common point.
\end{itemize}

\item[(2)] When points N and P in the coordinate space satisfy all of the following conditions (iii), (iv), (v), find the volume $W$ of the region that P can occupy. If necessary, you may use a real number $\alpha$ satisfying $\sin\alpha = \dfrac{1}{\sqrt{3}}$ $\left(0 < \alpha < \dfrac{\pi}{2}\right)$.

\begin{itemize}
\item[(iii)] $\mathrm{ON} + \mathrm{NP} \leq \sqrt{3}$
\item[(iv)] The line segment ON and $S$ have no common point.
\item[(v)] The line segment NP and $S$ have no common point, or have only the point P as a common point.
\end{itemize}

\end{enumerate}



%% Page 7
\textbf{1} \hfill \textit{Go to problem page}

\medskip

(1) For $A_k = \displaystyle\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} |\sin(x^2)|\,dx$, let $x^2 = t$ ($x = \sqrt{t}$), then $2x\,dx = dt$, so

$$A_k = \int_{k\pi}^{(k+1)\pi} |\sin t| \cdot \frac{dt}{2\sqrt{t}} = \int_{k\pi}^{(k+1)\pi} \frac{|\sin t|}{2\sqrt{t}}\,dt$$

Now, for $k\pi \leq t \leq (k+1)\pi$, we have $\dfrac{1}{\sqrt{(k+1)\pi}} \leq \dfrac{1}{\sqrt{t}} \leq \dfrac{1}{\sqrt{k\pi}}$, so

$$\frac{|\sin t|}{2\sqrt{(k+1)\pi}} \leq \frac{|\sin t|}{2\sqrt{t}} \leq \frac{|\sin t|}{2\sqrt{k\pi}}$$

Integrating each side from $k\pi$ to $(k+1)\pi$,

$$\frac{1}{2\sqrt{(k+1)\pi}}\int_{k\pi}^{(k+1)\pi} |\sin t|\,dt \leq A_k \leq \frac{1}{2\sqrt{k\pi}}\int_{k\pi}^{(k+1)\pi} |\sin t|\,dt \quad \cdots\cdots\cdots\textcircled{1}$$

Here, since $\sin t$ does not change sign for $k\pi \leq t \leq (k+1)\pi$,

$$\int_{k\pi}^{(k+1)\pi} |\sin t|\,dt = \left|\int_{k\pi}^{(k+1)\pi} \sin t\,dt\right| = \left|-[\cos t]_{k\pi}^{(k+1)\pi}\right|$$
$$= \left|-\cos(k+1)\pi + \cos k\pi\right| = \left|(-1)^{k+1} + (-1)^k\right| = \left|(-1)^k\right||-1+1| = 2$$

Therefore, from \textcircled{1},

$$\frac{1}{\sqrt{(k+1)\pi}} \leq A_k \leq \frac{1}{\sqrt{k\pi}} \quad \cdots\cdots\cdots\textcircled{2}$$

(2) $\displaystyle\int_{\sqrt{n\pi}}^{\sqrt{2n\pi}} |\sin(x^2)|\,dx = \sum_{k=n}^{2n-1}\left(\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} |\sin(x^2)|\,dx\right) = \sum_{k=n}^{2n-1} A_k$, so from \textcircled{2},

$$\sum_{k=n}^{2n-1} \frac{1}{\sqrt{(k+1)\pi}} \leq \sum_{k=n}^{2n-1} A_k \leq \sum_{k=n}^{2n-1} \frac{1}{\sqrt{k\pi}}$$

Since $B_n = \dfrac{1}{\sqrt{n}}\displaystyle\int_{\sqrt{n\pi}}^{\sqrt{2n\pi}} |\sin(x^2)|\,dx$,

$$\frac{1}{\sqrt{n}}\sum_{k=n}^{2n-1} \frac{1}{\sqrt{(k+1)\pi}} \leq B_n \leq \frac{1}{\sqrt{n}}\sum_{k=n}^{2n-1} \frac{1}{\sqrt{k\pi}} \quad \cdots\cdots\cdots\textcircled{3}$$

Here, letting $l = k - n$ and considering $n \to \infty$,

$$\frac{1}{\sqrt{n}}\sum_{k=n}^{2n-1} \frac{1}{\sqrt{k\pi}} = \frac{1}{\sqrt{n}}\sum_{l=0}^{n-1} \frac{1}{\sqrt{(n+l)\pi}} = \frac{1}{n}\sum_{l=0}^{n-1} \frac{1}{\sqrt{\left(1+\dfrac{l}{n}\right)\pi}}$$

$$\to \int_0^1 \frac{1}{\sqrt{(1+x)\pi}}\,dx \quad (n\to\infty)$$

Also, letting $m = k+1-n$ and considering $n \to \infty$,

$$\frac{1}{\sqrt{n}}\sum_{k=n}^{2n-1} \frac{1}{\sqrt{(k+1)\pi}} = \frac{1}{\sqrt{n}}\sum_{m=1}^{n} \frac{1}{\sqrt{(n+m)\pi}} = \frac{1}{n}\sum_{m=1}^{n} \frac{1}{\sqrt{\left(1+\dfrac{m}{n}\right)\pi}}$$

$$\to \int_0^1 \frac{1}{\sqrt{(1+x)\pi}}\,dx \quad (n\to\infty)$$



%% Page 8
Thus, from \textcircled{3},

$$\lim_{n \to \infty} B_n = \int_0^1 \frac{1}{\sqrt{(1+x)\pi}}\,dx = \frac{1}{\sqrt{\pi}}\bigl[2\sqrt{(1+x)}\bigr]_0^1 = \frac{2}{\sqrt{\pi}}(\sqrt{2}-1)$$

\textbf{[Commentary]}

This is a standard problem combining definite integrals and inequalities with the Riemann sum method. However, it has a somewhat formal atmosphere.



%% Page 9
\noindent\textbf{2} \hfill (Go to problem page)

\medskip

\noindent(1) When arranging all 12 balls --- 3 black balls, 4 red balls, and 5 white balls --- in a single row, assume that all $12!$ arrangements are equally likely.

\medskip

The probability that no two red balls are adjacent is found by first arranging the 8 white and black balls in a row, then choosing 4 of the 9 gaps (between balls or at either end) to insert the 4 red balls one by one. Thus the probability $p$ is,
$$p = \frac{8! \times {}_9P_4}{12!} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9} = \frac{14}{55}$$

\noindent(2) The probability that no two red balls are adjacent and no two black balls are adjacent is found by considering the following cases:

\medskip

\noindent(i) \textbf{When the arrangement red-black-black-red-black occurs}

\medskip

First arrange a group of 5 white balls and 3 black balls together, then insert 2 of the 4 red balls into the gaps between the black balls, and choose 2 of the 7 positions between the group of 5 white balls and 3 black balls or at either end to insert the remaining 2 red balls one by one. The probability is,
$$\frac{(3! \times 6!) \times {}_4P_2 \times {}_7P_2}{12!} = \frac{3 \cdot 2 \cdot 4 \cdot 3 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7} = \frac{1}{2^2 \cdot 5 \cdot 11}$$

\noindent(ii) \textbf{When the arrangement black-red-black occurs but the arrangement red-black-black-red-black does not occur}

\medskip

First arrange 5 white balls, then choose 2 of the 6 gaps (between balls or at either end) to place a group of 2 black balls and 1 single black ball. Next, insert 1 of the 4 red balls into the gap between the black balls, and choose 3 of the 8 positions between the 5 white balls, the group of 2 black balls, and the single black ball or at either end to insert the remaining 3 red balls one by one. The probability is,
$$\frac{5! \times ({}_3P_2 \times 1 \times {}_6P_2) \times {}_4P_1 \times {}_8P_3}{12!} = \frac{3 \cdot 2 \cdot 6 \cdot 5 \cdot 4 \cdot 8 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6} = \frac{2}{3 \cdot 11}$$

\noindent(iii) \textbf{When the arrangement black-red-black does not occur}

\medskip

First arrange 5 white balls, then choose 3 of the 6 gaps (between balls or at either end) to place the 3 black balls one by one. Next, choose 4 of the 9 gaps between the 5 white balls and 3 black balls or at either end to insert the 4 red balls one by one. The probability is,
$$\frac{5! \times {}_6P_3 \times {}_9P_4}{12!} = \frac{6 \cdot 5 \cdot 4 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6} = \frac{1}{11}$$

\noindent From (i)--(iii), letting $r$ denote the probability that no two red balls are adjacent and no two black balls are adjacent,



%% Page 10
$$r = \frac{1}{2^2 \cdot 5 \cdot 11} + \frac{2}{3 \cdot 11} + \frac{1}{11} = \frac{3 + 2 \cdot 20 + 60}{2^2 \cdot 5 \cdot 3 \cdot 11} = \frac{103}{2^2 \cdot 5 \cdot 3 \cdot 11}$$

Therefore, the conditional probability $q$ that no black balls are adjacent, given that no red balls are adjacent, is
$$q = \frac{r}{p} = \frac{103}{2^2 \cdot 5 \cdot 3 \cdot 11} \div \frac{14}{55} = \frac{103}{168}$$

\textbf{[Commentary]}

This is a standard probability problem. Part (1) involves routine processing, while part (2) admits various approaches. Here, we perform case analysis by focusing on the arrangement black-red-black.



%% Page 11
\section*{\boxed{3} \hfill \text{Go to Problem Page}}

\begin{enumerate}
\item[(1)] When any point $(\cos\theta,\ a+\sin\theta)$ on the circle $C: x^2+(y-a)^2=1$ is contained in the region represented by the inequality $y>x^2$, we have $a+\sin\theta > \cos^2\theta$, so

$$a > \cos^2\theta - \sin\theta = 1-\sin^2\theta - \sin\theta = -\!\left(\sin\theta+\frac{1}{2}\right)^2+\frac{5}{4} \quad \cdots\cdots\textcircled{1}$$

Here, since $\theta$ takes any value, we have $-1\leq\sin\theta\leq 1$, and the maximum value of the right-hand side of \textcircled{1} is $\dfrac{5}{4}$,

so the range of $a$ for which \textcircled{1} always holds is $a>\dfrac{5}{4}$.

\item[(2)] From (1), when $a>\dfrac{5}{4}$, the equation of the tangent line to $C$ at the point $\mathrm{P}(\cos\theta,\ a+\sin\theta)$ on $S$

$\left(-\dfrac{\pi}{2}\leq\theta<0\right)$ is

$$x\cos\theta+(a+\sin\theta-a)(y-a)=1$$
$$x\cos\theta+y\sin\theta = a\sin\theta+1 \quad\cdots\cdots\textcircled{2}$$

The intersection points with the parabola $y=x^2\cdots\cdots\textcircled{3}$ are found by solving \textcircled{2} and \textcircled{3} simultaneously:

$$x\cos\theta + x^2\sin\theta = a\sin\theta+1$$
$$x^2\sin\theta + x\cos\theta - (a\sin\theta+1)=0 \quad\cdots\cdots\textcircled{3}'$$

Since $-1\leq\sin\theta<0$, letting $D$ be the discriminant of \textcircled{3}$'$, when $a>\dfrac{5}{4}$,

$$D = \cos^2\theta + 4\sin\theta(a\sin\theta+1) = (4a-1)\sin^2\theta+4\sin\theta+1$$
$$=(4a-1)\!\left(\sin\theta+\frac{2}{4a-1}\right)^2+\frac{4a-5}{4a-1}>0$$

So, letting the real solutions of \textcircled{3}$'$ be $x=\alpha,\ \beta\ (\alpha<\beta)$,

$$\alpha = \frac{-\cos\theta+\sqrt{D}}{2\sin\theta},\quad \beta = \frac{-\cos\theta-\sqrt{D}}{2\sin\theta}$$

Now, the intersection points are expressed as $(\alpha,\ \alpha^2)$, $(\beta,\ \beta^2)$, and the length $L_{\mathrm{P}}$ of the chord cut off is

$$L_{\mathrm{P}}^2 = (\alpha-\beta)^2+(\alpha^2-\beta^2)^2 = (\alpha-\beta)^2\{1+(\alpha+\beta)^2\} = \frac{D}{\sin^2\theta}\!\left(1+\frac{\cos^2\theta}{\sin^2\theta}\right)$$

$$= \frac{D}{\sin^2\theta}\cdot\frac{1}{\sin^2\theta} = \frac{D}{\sin^4\theta} = \frac{(4a-1)\sin^2\theta+4\sin\theta+1}{\sin^4\theta}$$

Here, letting $u=\sin\theta\ (-1\leq u<0)$, $\theta$ and $u$ correspond one-to-one, and further

$$f(u) = \frac{(4a-1)u^2+4u+1}{u^4} = \frac{4a-1}{u^2}+\frac{4}{u^3}+\frac{1}{u^4}$$

gives $L_{\mathrm{P}}^2 = f(u)$, and

$$f'(u) = \frac{-2(4a-1)}{u^3}-\frac{12}{u^4}-\frac{4}{u^5} = -\frac{2}{u^5}\{(4a-1)u^2+6u+2\}$$

So, letting $g(u)=(4a-1)u^2+6u+2$, we have $f'(u)=-\dfrac{2}{u^5}g(u)$,

and noting that $-\dfrac{2}{u^5}>0$, the signs of $f'(u)$ and $g(u)$ agree,

\end{enumerate}

\begin{center}
$-5-$ \hfill \copyright\ 電送数学舎 2023
\end{center}



%% Page 12
$$g(u) = (4a-1)\left(u + \frac{3}{4a-1}\right)^2 + \frac{8a-11}{4a-1}$$

Since $a > \dfrac{5}{4}$, we have $g(-1) = (4a-1) - 6 + 2 = 4a - 5 > 0$, $g(0) = 2 > 0$, and $-\dfrac{3}{4} < -\dfrac{3}{4a-1} < 0$. Based on this, considering the sign of $g(u)$, i.e., $f'(u)$:

\begin{itemize}
\item[(i)] When $8a - 11 \geq 0$ $\left(a \geq \dfrac{11}{8}\right)$

$f'(u) \geq 0$ for $-1 \leq u < 0$, so $f(u)$ is monotonically increasing.

\item[(ii)] When $8a - 11 < 0$ $\left(\dfrac{5}{4} < a < \dfrac{11}{8}\right)$

$f'(u) = 0$ has two distinct real roots in $-1 \leq u < 0$; denoting these as $u = u_1,\ u_2\ (u_1 < u_2)$, the increase/decrease of $f(u)$ is as shown in the table on the right. From this, for a certain value of $f(u)$, there exist multiple values of $u$.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$u$ & $-1$ & $\cdots$ & $u_1$ & $\cdots$ & $u_2$ & $\cdots$ & $0$ \\
\hline
$f'(u)$ & & $+$ & $0$ & $-$ & $0$ & $+$ & \\
\hline
$f(u)$ & $4a-4$ & $\nearrow$ & & $\searrow$ & & $\nearrow$ & $\infty$ \\
\hline
\end{tabular}
\end{center}

\end{itemize}

From (i) and (ii), the condition for there to exist two distinct points Q, R on $S$ such that $L_{\mathrm{Q}} = L_{\mathrm{R}}$ is:
$$\frac{5}{4} < a < \frac{11}{8}$$

\bigskip

\noindent\textbf{[Commentary]}

This is an application problem of differentiation dealing with the relationship between a parabola and a line. In part (2), the amount of computation varies depending on the initial setup method, but in any case, it suffices to process things so that the graph of the corresponding function $L_{\mathrm{P}}^2 = f(u)$ is sinuous (has a local maximum and minimum).



%% Page 13
\noindent\textbf{4} \hfill (Go to problem page)

\medskip

\noindent(1) For the 4 points $O(0,0,0)$, $A(2,0,0)$, $B(1,1,1)$, $C(1,2,3)$, let $P(x,y,z)$. From $\overrightarrow{OP}\cdot\overrightarrow{OA}=0$, $\overrightarrow{OP}\cdot\overrightarrow{OB}=0$, $\overrightarrow{OP}\cdot\overrightarrow{OC}=1$,
$$2x=0,\quad x+y+z=0,\quad x+2y+3z=1$$
From these, $x=0$, $y=-1$, $z=1$, so $P(0,-1,1)$.

\medskip

\noindent(2) Let $\overrightarrow{OH}=(1-t)\overrightarrow{OA}+t\overrightarrow{OB}=(2-t,\,t,\,t)$, then
$$\overrightarrow{PH}=(2-t,\,t,\,t)-(0,-1,1)$$
$$=(2-t,\;t+1,\;t-1)$$
Also, $\overrightarrow{AB}=(-1,1,1)$, and from $\overrightarrow{PH}\cdot\overrightarrow{AB}=0$,
$$-(2-t)+(t+1)+(t-1)=0$$
Thus $3t-2=0$, so $t=\dfrac{2}{3}$, and $\overrightarrow{OH}=\dfrac{1}{3}\overrightarrow{OA}+\dfrac{2}{3}\overrightarrow{OB}$.

\hfill\textit{[Figure: Points P, C, B, O, H, A in 3D]}

\medskip

\noindent(3) From $\overrightarrow{OP}\perp\overrightarrow{OA}$, $\overrightarrow{OP}\perp\overrightarrow{OB}$, OP is perpendicular to the plane OAB, and furthermore from $\overrightarrow{PH}\perp\overrightarrow{AB}$, $\overrightarrow{OH}\perp\overrightarrow{AB}$.

Now, let $\overrightarrow{OR}=\dfrac{3}{4}\overrightarrow{OA}$, and consider the sphere $S$ of radius $r$ centered at the point $Q$ defined by $\overrightarrow{OQ}=\overrightarrow{OR}+\overrightarrow{OP}$, and find the range of $r$ for which $S$ has a common point with $\triangle OHB$.

\hfill\textit{[Figure: Points P, Q, O, B, S, H, R, A in 3D]}

First, drop a perpendicular from R to OH and let the foot be S; then the minimum value of $r$ is QS.

Here, $\mathrm{QR}=\mathrm{PO}=\sqrt{(-1)^2+1^2}=\sqrt{2}$, $\mathrm{AB}=\sqrt{(-1)^2+1^2+1^2}=\sqrt{3}$, and
$$\mathrm{RS}:\mathrm{AH}=\mathrm{OR}:\mathrm{OA}=3:4,\quad \mathrm{AH}:\mathrm{AB}=2:3$$
From these, $\mathrm{RS}=\dfrac{3}{4}\mathrm{AH}=\dfrac{3}{4}\cdot\dfrac{2}{3}\mathrm{AB}=\dfrac{\sqrt{3}}{2}$, and since QR is perpendicular to the plane ABC,
$$\mathrm{QS}=\sqrt{\mathrm{QR}^2+\mathrm{RS}^2}=\sqrt{(\sqrt{2})^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\frac{\sqrt{11}}{2}$$

Also, from $R\!\left(\dfrac{3}{2},\,0,\,0\right)$, $\mathrm{RO}=\dfrac{3}{2}$, $\mathrm{RB}=\sqrt{\left(\dfrac{1}{2}\right)^2+(-1)^2+(-1)^2}=\dfrac{3}{2}$,

and since $\mathrm{RH}<\mathrm{RB}$, the maximum value of $r$ is QO or QB, and
$$\mathrm{QO}=\sqrt{\mathrm{QR}^2+\mathrm{OR}^2}=\sqrt{(\sqrt{2})^2+\left(\frac{3}{2}\right)^2}=\frac{\sqrt{17}}{2}$$

From the above, the required range of $r$ is $\dfrac{\sqrt{11}}{2}\leq r\leq\dfrac{\sqrt{17}}{2}$.

\bigskip

\noindent\textbf{[Commentary]}

This is a standard problem on applications of vectors in space. By noting that $\triangle OHB$ is a right triangle, the rest reduces to computation.



%% Page 14
\section*{\boxed{5} \hfill \text{Go to Problem Page}}

\textbf{(1)} Let $q_1(x)$ be the quotient and $r(x)$ be the remainder when $g(x)$ is divided by $f(x)$, so that
$$g(x) = f(x)q_1(x) + r(x)$$
Raising both sides to the 7th power,
$$g(x)^7 = \{f(x)q_1(x)+r(x)\}^7 = \sum_{k=0}^{6} {}_7C_k\, f(x)^{7-k}\,q_1(x)^{7-k}\,r(x)^k + r(x)^7$$
$$= f(x)\sum_{k=0}^{6} {}_7C_k\, f(x)^{6-k}\,q_1(x)^{7-k}\,r(x)^k + r(x)^7$$
From this, $g(x)^7 - r(x)^7 = f(x)\displaystyle\sum_{k=0}^{6} {}_7C_k\, f(x)^{6-k}\,q_1(x)^{7-k}\,r(x)^k$, so $g(x)^7 - r(x)^7$ is divisible by $f(x)$.

That is, the remainder when $g(x)^7$ is divided by $f(x)$ equals the remainder when $r(x)^7$ is divided by $f(x)$.

\textbf{(2)} Let $q_2(x)$ be the quotient and $h_1(x)$ be the remainder when $h(x)^7$ is divided by $f(x)$, and let $q_3(x)$ be the quotient and $h_2(x)$ be the remainder when $h_1(x)^7$ is divided by $f(x)$. If $h_2(x) = h(x)$, then
$$h(x)^7 = f(x)q_2(x) + h_1(x) \cdots\cdots\textcircled{1}, \quad h_1(x)^7 = f(x)q_3(x) + h(x) \cdots\cdots\textcircled{2}$$
From \textcircled{1}\textcircled{2}, $\{h(x)^7 - f(x)q_2(x)\}^7 = f(x)q_3(x) + h(x)$.

By the same argument as (1), $h(x)^{49} - h(x)$ is divisible by $f(x) = (x-1)^2(x-2)$.

Now, let $H(x) = h(x)^{49} - h(x)$. We show that the necessary and sufficient condition for the polynomial $H(x)$ to be divisible by $f(x)$ is that $H(1)=0\cdots\cdots\textcircled{3}$, $H(2)=0\cdots\cdots\textcircled{4}$, $H'(1)=0\cdots\cdots\textcircled{5}$.

\textbf{(a)} Let $Q_1(x)$ be the quotient when $H(x)$ is divided by $f(x)$:
$$H(x) = (x-1)^2(x-2)Q_1(x)$$
$$H'(x) = 2(x-1)(x-2)Q_1(x) + (x-1)^2 Q_1(x) + (x-1)^2(x-2)Q_1'(x)$$
From this, \textcircled{3}\textcircled{4}\textcircled{5} hold.

\textbf{(b)} Conversely, when \textcircled{3}\textcircled{4}\textcircled{5} hold, from \textcircled{3}\textcircled{4}, $H(x)$ is divisible by $(x-1)(x-2)$.

Let $Q_2(x)$ be the quotient when $H(x)$ is divided by $(x-1)(x-2)$:
$$H(x) = (x-1)(x-2)Q_2(x)$$
$$H'(x) = (x-2)Q_2(x) + (x-1)Q_2(x) + (x-1)(x-2)Q_2'(x)$$
Then from \textcircled{5}, $Q_2(1) = 0$, so $Q_2(x)$ is divisible by $x-1$. Let $Q_3(x)$ be the quotient when $Q_2(x)$ is divided by $x-1$:
$$Q_2(x) = (x-1)Q_3(x)$$
Putting it together, $H(x) = (x-1)(x-2)(x-1)Q_3(x) = (x-1)^2(x-2)Q_3(x)$, so $H(x)$ is divisible by $f(x)$.

From (a)(b), the necessary and sufficient condition for $H(x)$ to be divisible by $f(x)$ is \textcircled{3} and \textcircled{4} and \textcircled{5}.

Now, from $h(x) = x^2 + ax + b$, we have $H(x) = (x^2+ax+b)^{49} - (x^2+ax+b)$.



%% Page 15
$$H'(x) = 49(x^2+ax+b)^{48}(2x+a)-(2x+a)$$

From \textcircled{3}, $(1+a+b)^{49}-(1+a+b)=0$, so
$$(1+a+b)^{49}=1+a+b \cdots\cdots\textcircled{6}$$

From \textcircled{4}, $(4+2a+b)^{49}-(4+2a+b)=0$, so
$$(4+2a+b)^{49}=4+2a+b \cdots\cdots\textcircled{7}$$

From \textcircled{5}, $49(1+a+b)^{48}(2+a)-(2+a)=0$, so
$$49(1+a+b)^{48}(2+a)=2+a \cdots\cdots\textcircled{8}$$

\noindent(i) When $1+a+b=0$

\textcircled{6} holds, and from \textcircled{8}, $2+a=0$, i.e., $a=-2$, so $b=-1-(-2)=1$.

In this case, $4+2a+b=1$, and \textcircled{7} holds.

\noindent(ii) When $1+a+b\neq 0$

From \textcircled{6}, $(1+a+b)^{48}=1$, so $1+a+b=\pm 1$.

\noindent(ii-i) When $1+a+b=1$

From \textcircled{8}, $49(2+a)=2+a$, so $2+a=0$, i.e., $a=-2$.

Also, $b=1+2-1=2$, and since $4+2a+b=2$, \textcircled{7} does not hold.

\noindent(ii-ii) When $1+a+b=-1$

From \textcircled{8}, $49(2+a)=2+a$, so $2+a=0$, i.e., $a=-2$.

Also, $b=-1+2-1=0$, and since $4+2a+b=0$, \textcircled{7} holds.

From (i)(ii), the pairs $(a,\,b)$ sought are $(a,\,b)=(-2,\,1),\,(-2,\,0)$.

\bigskip
\noindent\textbf{[Commentary]}

This is a problem based on polynomial division. In (2), by setting $g(x)=h(x)^7$ and applying the result of (1), the solution becomes somewhat more concise. Also, regarding the description that the necessary and sufficient condition for $H(x)$ to be divisible by $f(x)$ is \textcircled{3} and \textcircled{4} and \textcircled{5}, we were uncertain whether it could be omitted. In the solution above, we have noted it briefly.



%% Page 16
\noindent\textbf{6} \hfill \textit{Go to problem page}

\medskip

\noindent(1) In a coordinate space with O as the origin, let $\mathrm{A}(1,1,1)$, $\mathrm{B}(-1,1,1)$, $\mathrm{C}(-1,-1,1)$, $\mathrm{D}(1,-1,1)$, $\mathrm{E}(1,1,-1)$, $\mathrm{F}(-1,1,-1)$, $\mathrm{G}(-1,-1,-1)$, $\mathrm{H}(1,-1,-1)$.

\medskip

\noindent The region $S$ satisfying $z < 1$ among the surface of the cube defined by $|x|\leq 1$, $|y|\leq 1$, $|z|\leq 1$ represents the five faces of cube ABCD-EFGH other than face ABCD.

\medskip

\noindent Now, the range $V$ of points P satisfying $\mathrm{OP} \leq \sqrt{3}$ and the condition that segment OP and $S$ share no common point or share only point P is:

\medskip

\noindent\textbf{(a)} The interior or surface of cube ABCD-EFGH

\noindent\hspace{1em} In this case, the volume of the range of point P is $2^3 = 8$.

\medskip

\noindent\textbf{(b)} The upper side of plane ABCD

\noindent\hspace{1em} In this case, the range of point P is the upper side of planes OAB, OBC, OCD, ODA, and the interior or surface of a sphere of radius $\sqrt{3}$ centered at O. By symmetry, its volume is
$$\frac{1}{6}\left\{\frac{4}{3}\pi(\sqrt{3})^3 - 2^3\right\} = \frac{2}{3}\sqrt{3}\pi - \frac{4}{3}.$$

\noindent From (a) and (b), the volume of range $V$ is
$$8 + \left(\frac{2}{3}\sqrt{3}\pi - \frac{4}{3}\right) = \frac{2}{3}\sqrt{3}\pi + \frac{20}{3}.$$

\medskip

\noindent(2) The range $W$ of points P satisfying $\mathrm{ON} + \mathrm{NP} \leq \sqrt{3}$, with segment ON and $S$ sharing no common point, and segment NP and $S$ sharing no common point or sharing only point P is:

\medskip

\noindent\textbf{(c)} When O, N, P are collinear

\noindent\hspace{1em} The range of point P in this case is the same as in (1), so its volume is $\dfrac{2}{3}\sqrt{3}\pi + \dfrac{20}{3}$.

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\noindent\textbf{(d)} When O, N, P are not collinear

\noindent\hspace{1em} Below, we find the range of point P not included in case (c).

\noindent\hspace{1em} In this case, it suffices to consider when point N is on a side of square ABCD, so first let M be the midpoint of side AB and consider the case where point N is on segment AM.

\noindent\hspace{1em} First, as preparation, expressing the range of point P from case (b):
$$z \geq 1,\quad z \geq x,\quad z \geq -x,\quad z \geq y,\quad z \geq -y,\quad x^2+y^2+z^2 \leq 3$$

\noindent\hspace{1em} Here, letting $0 \leq t \leq 1$, the inequalities representing the cross-section at the plane $x = t$ passing through point $\mathrm{N}(t, 1, 1)$ are:
$$z \geq 1,\quad z \geq t,\quad z \geq -t,\quad z \geq y,\quad z \geq -y$$
$$y^2 + z^2 \leq 3 - t^2$$

\noindent\hspace{1em} This is shown as the shaded region in the figure on the right.

\hfill \textit{[Figure: cross-section in the $yz$-plane showing shaded region bounded by $z\geq 1$, $z\geq \pm y$, and $y^2+z^2\leq 3-t^2$, with $\sqrt{3-t^2}$ marked on the $z$-axis and N marked]}

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$-10-$ \hfill \copyright\ 電送数学舎 2023
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When O, N, P are not collinear, the range of the newly added point P forms a sector centered at N with radius $r = \sqrt{3-t^2} - \sqrt{2}$ and central angle $\dfrac{3}{4}\pi$. As shown in the figure, this corresponds to the darkly shaded region in the figure on the right.

\textit{[Figure: A sector in the $yz$-plane centered at N with radius $\sqrt{3-t^2}$, with shaded region]}

Note that when $t = 0 \to 1$, $r = \sqrt{3} - \sqrt{2} \to 0$, and $r$ decreases monotonically as $t$ increases.

Now, this cross-sectional area is $\dfrac{1}{2}\!\left(\sqrt{3-t^2}-\sqrt{2}\right)^2 \cdot \dfrac{3}{4}\pi = \dfrac{3}{8}\pi\!\left(5 - t^2 - 2\sqrt{2}\sqrt{3-t^2}\right)$,

and considering symmetry, the volume of the range of the newly added point P when O, N, P are not collinear is:

$$4 \times 2\int_0^1 \frac{3}{8}\pi\!\left(5 - t^2 - 2\sqrt{2}\sqrt{3-t^2}\right)dt = 3\pi\int_0^1\!\left(5 - t^2 - 2\sqrt{2}\sqrt{3-t^2}\right)dt$$

$$= 3\pi\!\left[5t - \frac{t^3}{3}\right]_0^1 - 6\sqrt{2}\pi\int_0^1\sqrt{3-t^2}\,dt = 14\pi - 6\sqrt{2}\pi\int_0^1\sqrt{3-t^2}\,dt$$

Here, let $I = \displaystyle\int_0^1 \sqrt{3-t^2}\,dt$, and substitute $t = \sqrt{3}\sin\theta$ $\left(-\dfrac{\pi}{2} \leqq \theta \leqq \dfrac{\pi}{2}\right)$,

then $dt = \sqrt{3}\cos\theta\,d\theta$, and since $\sin\alpha = \dfrac{1}{\sqrt{3}}$, when $t = 0 \to 1$, $\theta = 0 \to \alpha$.

$$I = \int_0^{\alpha}\sqrt{3 - 3\sin^2\theta}\cdot\sqrt{3}\cos\theta\,d\theta = 3\int_0^{\alpha}\cos^2\theta\,d\theta = \frac{3}{2}\int_0^{\alpha}(1+\cos 2\theta)\,d\theta$$

$$= \frac{3}{2}\!\left[\theta + \frac{1}{2}\sin 2\theta\right]_0^{\alpha} = \frac{3}{2}\alpha + \frac{3}{4}\sin 2\alpha = \frac{3}{2}\alpha + \frac{3}{2}\sin\alpha\cos\alpha$$

$$= \frac{3}{2}\alpha + \frac{3}{2}\cdot\frac{1}{\sqrt{3}}\sqrt{1 - \frac{1}{3}} = \frac{3}{2}\alpha + \frac{\sqrt{2}}{2}$$

Therefore, $14\pi - 6\sqrt{2}\pi I = 14\pi - 6\sqrt{2}\pi\!\left(\dfrac{3}{2}\alpha + \dfrac{\sqrt{2}}{2}\right) = (8 - 9\sqrt{2}\alpha)\pi$.

From (c)(d), the volume of the region $W$ is:

$$\left(\frac{2}{3}\sqrt{3}\pi + \frac{20}{3}\right) + (8 - 9\sqrt{2}\alpha)\pi = \left(8 + \frac{2}{3}\sqrt{3} - 9\sqrt{2}\alpha\right)\pi + \frac{20}{3}$$

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\noindent\textbf{[Commentary]}

This is a difficult problem asking for the volume of a solid. For (1), I initially tried to compute it by integration but it did not work out well, so I used a method utilizing symmetry. For (2), since the segment OP bends, the problem asks for the volume of the region ``overflowing'' from each side of the square ABCD on the top face of the cube, but the description relies considerably on intuition.