\noindent\textbf{6}

\medskip

An integer greater than or equal to 2 that has no positive divisors other than 1 and itself is called a prime number. Answer the following questions.

\medskip

\noindent(1) Let $f(x) = x^3 + 10x^2 + 20x$. Find all integers $n$ such that $f(n)$ is a prime number.

\medskip

\noindent(2) Let $a$, $b$ be integer constants, and let $g(x) = x^3 + ax^2 + bx$. Show that the number of integers $n$ such that $g(n)$ is a prime number is at most 3.



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

For a point $\mathrm{P}(x,\ y,\ 0)$ in the $xy$-plane different from the origin O, and point $\mathrm{A}(0,\ -1,\ 1)$,

$$\cos\angle\mathrm{AOP} = \frac{\overrightarrow{\mathrm{OA}}\cdot\overrightarrow{\mathrm{OP}}}{|\overrightarrow{\mathrm{OA}}||\overrightarrow{\mathrm{OP}}|} = \frac{-y}{\sqrt{2}\sqrt{x^2+y^2}}$$

Since $\angle\mathrm{AOP} \geq \dfrac{2}{3}\pi$ implies $\cos\angle\mathrm{AOP} \leq -\dfrac{1}{2}$, we have $\dfrac{-y}{\sqrt{2}\sqrt{x^2+y^2}} \leq -\dfrac{1}{2}$

$$2y \geq \sqrt{2}\sqrt{x^2+y^2}$$

Then, for $y \geq 0$, we get $4y^2 \geq 2(x^2+y^2)$, and from $y^2 - x^2 \geq 0$,
$$(y+x)(y-x) \geq 0 \quad (y \geq 0) \cdots\cdots\textcircled{1}$$

Also, from $\overrightarrow{\mathrm{AO}} = (0,\ 1,\ -1)$, $\overrightarrow{\mathrm{AP}} = (x,\ y+1,\ -1)$,

$$\cos\angle\mathrm{OAP} = \frac{\overrightarrow{\mathrm{AO}}\cdot\overrightarrow{\mathrm{AP}}}{|\overrightarrow{\mathrm{AO}}||\overrightarrow{\mathrm{AP}}|} = \frac{y+2}{\sqrt{2}\sqrt{x^2+(y+1)^2+1}}$$

Since $\angle\mathrm{OAP} \leq \dfrac{\pi}{6}$ implies $\cos\angle\mathrm{OAP} \geq \dfrac{\sqrt{3}}{2}$, we have $\dfrac{y+2}{\sqrt{2}\sqrt{x^2+(y+1)^2+1}} \geq \dfrac{\sqrt{3}}{2}$

$$2(y+2) \geq \sqrt{6}\sqrt{x^2+(y+1)^2+1}$$

Then, for $y+2 \geq 0$, we get $4(y+2)^2 \geq 6\{x^2+(y+1)^2+1\}$,
$$2(y^2+4y+4) \geq 3x^2+3(y^2+2y+1)+3$$
From $3x^2+y^2-2y-2 \leq 0$, we get $3x^2+(y-1)^2 \leq 3$, and thus

$$x^2 + \frac{(y-1)^2}{3} \leq 1 \quad (y \geq -2) \cdots\cdots\textcircled{2}$$

From \textcircled{1} and \textcircled{2}, the region in the $xy$-plane that P can occupy is the shaded region in the figure on the right. The boundary except for the origin is included.

\textit{[Figure: Shaded region in the $xy$-plane bounded by the ellipse $x^2+\frac{(y-1)^2}{3}\leq 1$ and the lines $y=x$, $y=-x$ with $y\geq 0$; key points marked at $y=1+\sqrt{3}$, $y=1$, $y=1-\sqrt{3}$, and $x=\pm 1$]}

\subsection*{[Commentary]}

This is a basic problem on spatial vectors and regions. The amount of computation is relatively small.



%% Page 8
\textbf{2} \hfill \textit{Go to problem page}

\medskip

(1) For $f(x) = \displaystyle\int_0^1 \frac{|t-x|}{1+t^2}\,dt\ (0 \leqq x \leqq 1)$, we have $f(x) = -\displaystyle\int_0^x \frac{t-x}{1+t^2}\,dt + \int_x^1 \frac{t-x}{1+t^2}\,dt$

$$f(x) = -\int_0^x \frac{t}{1+t^2}\,dt + x\int_0^x \frac{dt}{1+t^2} + \int_x^1 \frac{t}{1+t^2}\,dt - x\int_x^1 \frac{dt}{1+t^2} \quad \cdots\cdots\text{\textcircled{1}}$$

$$f'(x) = -\frac{x}{1+x^2} + \int_0^x \frac{dt}{1+t^2} + x\cdot\frac{1}{1+x^2} - \frac{x}{1+x^2} - \int_x^1 \frac{dt}{1+t^2} + x\cdot\frac{1}{1+x^2}$$

$$= \int_0^x \frac{dt}{1+t^2} - \int_x^1 \frac{dt}{1+t^2} \quad \cdots\cdots\text{\textcircled{2}}$$

From $0 < \alpha < \dfrac{\pi}{4}$, we have $0 < \tan\alpha < 1$, and $f'(\tan\alpha) = \displaystyle\int_0^{\tan\alpha} \frac{dt}{1+t^2} - \int_{\tan\alpha}^1 \frac{dt}{1+t^2}$.

Setting $t = \tan\theta\ \left(-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}\right)$, we get $dt = \dfrac{d\theta}{\cos^2\theta}$, so

$$f'(\tan\alpha) = \int_0^{\alpha} \frac{1}{1+\tan^2\theta}\cdot\frac{d\theta}{\cos^2\theta} - \int_{\alpha}^{\frac{\pi}{4}} \frac{1}{1+\tan^2\theta}\cdot\frac{d\theta}{\cos^2\theta}$$

$$= \int_0^{\alpha} d\theta - \int_{\alpha}^{\frac{\pi}{4}} d\theta = \alpha - \left(\frac{\pi}{4} - \alpha\right) = 2\alpha - \frac{\pi}{4}$$

Then, from $f'(\tan\alpha) = 0$, we get $2\alpha - \dfrac{\pi}{4} = 0$, so $\alpha = \dfrac{\pi}{8}$.

\medskip

(2) From the half-angle formula,
$$\tan^2\alpha = \tan^2\frac{\pi}{8} = \frac{1-\cos\dfrac{\pi}{4}}{1+\cos\dfrac{\pi}{4}} = \frac{1 - \dfrac{1}{\sqrt{2}}}{1 + \dfrac{1}{\sqrt{2}}} = \frac{\sqrt{2}-1}{\sqrt{2}+1} = (\sqrt{2}-1)^2$$

Since $\tan\dfrac{\pi}{8} > 0$, we have $\tan\alpha = \tan\dfrac{\pi}{8} = \sqrt{2}-1$.

\medskip

(3) From \textcircled{2}, $f''(x) = \dfrac{1}{1+x^2} + \dfrac{1}{1+x^2} = \dfrac{2}{1+x^2} > 0$, so $f'(x)$ is strictly increasing on $0 \leqq x \leqq 1$.

Then, from (2), since $f'(\sqrt{2}-1) = 0$, the increase/decrease of $f(x)$ on $0 \leqq x \leqq 1$ is as shown in the table on the right.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & $0$ & $\cdots$ & $\sqrt{2}-1$ & $\cdots$ & $1$ \\
\hline
$f'(x)$ & & $-$ & $0$ & $+$ & \\
\hline
$f(x)$ & & $\searrow$ & & $\nearrow$ & \\
\hline
\end{tabular}
\end{center}

At this point, from \textcircled{1},

$$f(0) = \int_0^1 \frac{t}{1+t^2}\,dt = \frac{1}{2}\bigl[\log(1+t^2)\bigr]_0^1 = \frac{1}{2}\log 2$$

$$f(1) = -\int_0^1 \frac{t}{1+t^2}\,dt + \int_0^1 \frac{dt}{1+t^2} = -\frac{1}{2}\log 2 + \int_0^{\frac{\pi}{4}} d\theta = \frac{\pi}{4} - \frac{1}{2}\log 2$$

Then, since $0.69 < \log 2 < 0.7$, we get $f(0) - f(1) = \log 2 - \dfrac{\pi}{4} < 0$, so $f(0) < f(1)$.

Therefore, the maximum value is $f(1) = \dfrac{\pi}{4} - \dfrac{1}{2}\log 2$.

Also, the minimum value is $f(\sqrt{2}-1)$, and from \textcircled{1},



%% Page 9
$$f(\sqrt{2}-1) = -\int_0^{\sqrt{2}-1}\frac{t}{1+t^2}dt + (\sqrt{2}-1)\int_0^{\sqrt{2}-1}\frac{dt}{1+t^2} + \int_{\sqrt{2}-1}^{1}\frac{t}{1+t^2}dt$$
$$-(\sqrt{2}-1)\int_{\sqrt{2}-1}^{1}\frac{dt}{1+t^2}$$

Here, $\displaystyle\int_0^{\sqrt{2}-1}\frac{t}{1+t^2}dt = \frac{1}{2}\bigl[\log(1+t^2)\bigr]_0^{\sqrt{2}-1} = \frac{1}{2}\log(4-2\sqrt{2})$

$$\int_{\sqrt{2}-1}^{1}\frac{t}{1+t^2}dt = \frac{1}{2}\log 2 - \frac{1}{2}\log(4-2\sqrt{2})$$

$$\int_0^{\sqrt{2}-1}\frac{dt}{1+t^2} - \int_{\sqrt{2}-1}^{1}\frac{dt}{1+t^2} = \int_0^{\frac{\pi}{8}}d\theta - \int_{\frac{\pi}{8}}^{\frac{\pi}{4}}d\theta = \frac{\pi}{8} - \left(\frac{\pi}{4}-\frac{\pi}{8}\right) = 0$$

Summarizing, $f(\sqrt{2}-1) = -\dfrac{1}{2}\log(4-2\sqrt{2}) + \dfrac{1}{2}\log 2 - \dfrac{1}{2}\log(4-2\sqrt{2})$, and

$$f(\sqrt{2}-1) = \log\sqrt{2} - \log(4-2\sqrt{2}) = \log\frac{\sqrt{2}}{4-2\sqrt{2}} = \log\frac{1}{2\sqrt{2}-2} = \log\frac{\sqrt{2}+1}{2}$$

\bigskip

\noindent\textbf{[Commentary]}

This is a computation problem involving definite integrals using the relationship between differentiation and integration. Part (1) serves as a guide for part (3). However, the computation of the minimum value is somewhat involved.



%% Page 10
\section*{\boxed{3} \hfill \text{Go to Problem Page}}

\begin{enumerate}
\item[(1)] The points symmetric to the point $(a,\,b)$ with respect to the $x$-axis, $y$-axis, the line $y=x$, and the line $y=-x$ are $(a,\,-b)$, $(-a,\,b)$, $(b,\,a)$, and $(-b,\,-a)$, respectively.

From this, the 8 points that point P, initially at $(2,\,1)$, can reach by the given rules are as follows.

$$\text{A}(2,\,1),\quad \text{B}(2,\,-1),\quad \text{C}(-2,\,1),\quad \text{D}(-2,\,-1)$$
$$\text{E}(1,\,2),\quad \text{F}(1,\,-2),\quad \text{G}(-1,\,2),\quad \text{H}(-1,\,-2)$$

\textit{[Figure: Coordinate plane showing points A through H at the described positions]}

\item[(2)] Let the probabilities that P is at A, B, C, D, E, F, G, H after $n$ seconds be $a_n$, $b_n$, $c_n$, $d_n$, $e_n$, $f_n$, $g_n$, $h_n$, respectively. Then,
$$a_{n+1} = \frac{1}{3}b_n + \frac{1}{3}c_n + \frac{1}{6}e_n + \frac{1}{6}h_n, \quad d_{n+1} = \frac{1}{3}c_n + \frac{1}{3}b_n + \frac{1}{6}h_n + \frac{1}{6}e_n$$

From this, $a_{n+1} = d_{n+1}$, so for $n \geq 2$, $a_n = d_n$.

Furthermore, since $a_1 = d_1 = 0$, we have $a_n = d_n$ $(n \geq 1)$.

\item[(3)] Similarly to (2), $a_n = d_n$, $b_n = c_n$, $e_n = h_n$, $f_n = g_n$, and
$$a_n + b_n + e_n + f_n = \frac{1}{2}$$

where $a_1 = 0$, $b_1 = \dfrac{1}{3}$, $e_1 = \dfrac{1}{6}$, $f_1 = 0$. In this case,

$$a_{n+1} = \frac{2}{3}b_n + \frac{1}{3}e_n \cdots\cdots\cdots\textcircled{1}, \quad b_{n+1} = \frac{2}{3}a_n + \frac{1}{3}f_n \cdots\cdots\cdots\textcircled{2}$$

$$e_{n+1} = \frac{2}{3}f_n + \frac{1}{3}a_n \cdots\cdots\cdots\textcircled{3}, \quad f_{n+1} = \frac{2}{3}e_n + \frac{1}{3}b_n \cdots\cdots\cdots\textcircled{4}$$

From $\textcircled{1}+\textcircled{4}$, $a_{n+1} + f_{n+1} = b_n + e_n = \dfrac{1}{2} - (a_n + f_n)$, so

$$a_{n+1} + f_{n+1} - \frac{1}{4} = -\left(a_n + f_n - \frac{1}{4}\right)$$

Thus, $a_n + f_n - \dfrac{1}{4} = \left(a_1 + f_1 - \dfrac{1}{4}\right)(-1)^{n-1} = -\dfrac{1}{4}(-1)^{n-1}$, so

$$a_n + f_n = \frac{1}{4}\{1-(-1)^{n-1}\} = \frac{1}{4}\{1+(-1)^n\} \cdots\cdots\cdots\textcircled{5}$$

From $\textcircled{1}-\textcircled{4}$, $a_{n+1} - f_{n+1} = \dfrac{1}{3}(b_n - e_n)$; from $\textcircled{2}-\textcircled{3}$, $b_{n+1} - e_{n+1} = \dfrac{1}{3}(a_n - f_n)$, so

$$a_{n+2} - f_{n+2} = \frac{1}{3}(b_{n+1} - e_{n+1}) = \frac{1}{9}(a_n - f_n) \cdots\cdots\cdots\textcircled{6}$$

Note that $a_2 = \dfrac{2}{3}b_1 + \dfrac{1}{3}e_1 = \dfrac{2}{9} + \dfrac{1}{18} = \dfrac{5}{18}$, $f_2 = \dfrac{2}{3}e_1 + \dfrac{1}{3}b_1 = \dfrac{1}{9} + \dfrac{1}{9} = \dfrac{2}{9}$.

From $\textcircled{6}$, $(a_{n+2} - f_{n+2}) - \dfrac{1}{3}(a_{n+1} - f_{n+1}) = -\dfrac{1}{3}\left\{(a_{n+1} - f_{n+1}) - \dfrac{1}{3}(a_n - f_n)\right\}$, so

$$(a_{n+1} - f_{n+1}) - \frac{1}{3}(a_n - f_n) = \left\{(a_2 - f_2) - \frac{1}{3}(a_1 - f_1)\right\}\left(-\frac{1}{3}\right)^{n-1} = \frac{1}{18}\left(-\frac{1}{3}\right)^{n-1}$$

From $\textcircled{6}$, $(a_{n+2} - f_{n+2}) + \dfrac{1}{3}(a_{n+1} - f_{n+1}) = \dfrac{1}{3}\left\{(a_{n+1} - f_{n+1}) + \dfrac{1}{3}(a_n - f_n)\right\}$, so

\end{enumerate}

\begin{center}
$-4-$ \hfill \copyright\ 電送数学舎 2024
\end{center}



%% Page 11
\begin{center}
2024 Tokyo University (Science) First Semester Exam \quad Solutions and Explanations
\end{center}

$$
(a_{n+1} - f_{n+1}) + \frac{1}{3}(a_n - f_n) = \left\{(a_2 - f_2) + \frac{1}{3}(a_1 - f_1)\right\}\left(\frac{1}{3}\right)^{n-1} = \frac{1}{18}\left(\frac{1}{3}\right)^{n-1}
$$

Therefore, $\dfrac{2}{3}(a_n - f_n) = \dfrac{1}{18}\left(\dfrac{1}{3}\right)^{n-1} - \dfrac{1}{18}\left(-\dfrac{1}{3}\right)^{n-1}$ and so,

$$
a_n - f_n = \frac{1}{12}\left(\frac{1}{3}\right)^{n-1} - \frac{1}{12}\left(-\frac{1}{3}\right)^{n-1} = \frac{1}{4}\left\{\left(\frac{1}{3}\right)^n + \left(-\frac{1}{3}\right)^n\right\} \cdots\cdots\cdots\textcircled{7}
$$

From $\textcircled{5} + \textcircled{7}$, $\;2a_n = \dfrac{1}{4}\left\{1 + (-1)^n + \left(\dfrac{1}{3}\right)^n + \left(-\dfrac{1}{3}\right)^n\right\}$, and thus,

$$
a_n = \frac{1}{8}\left\{1 + (-1)^n + \left(\frac{1}{3}\right)^n + \left(-\frac{1}{3}\right)^n\right\}
$$

From the above, the probability $a_n$ that P is at the point $(2,\,1)$ after $n$ seconds from the start is:

$$
a_n = \frac{1}{8}\left\{1 + 1 + \left(\frac{1}{3}\right)^n + \left(\frac{1}{3}\right)^n\right\} = \frac{1}{4}\left\{1 + \left(\frac{1}{3}\right)^n\right\} \quad (n \text{ is even})
$$

$$
a_n = \frac{1}{8}\left\{1 - 1 + \left(\frac{1}{3}\right)^n - \left(\frac{1}{3}\right)^n\right\} = 0 \quad (n \text{ is odd})
$$

\bigskip

\noindent\textbf{[Commentary]}

This is a problem on probability and recurrence relations. For the recurrence relations (1) through (4) in part (3), various solution methods can be considered. Note that from equation \textcircled{5}, it is also fine to split $n$ into even and odd cases for processing.



%% Page 12
\textbf{4} \hfill \textit{Go to problem page}

(1) For $f(x) = -\dfrac{\sqrt{2}}{4}x^2 + 4\sqrt{2} = -\dfrac{\sqrt{2}}{4}(x^2 - 16)$, the parabola $y = f(x)$ and the circle $C_t$ with center $\mathrm{C}(c(t),\ 0)$ and radius $r(t)$ share a common tangent at the point $\mathrm{P}(t,\ f(t))$ $(0 < t < 4)$.

Now, since $f'(x) = -\dfrac{\sqrt{2}}{2}x$, the direction vector $\vec{u}$ of the tangent to the parabola at point P can be written as $\vec{u} = \left(1,\ -\dfrac{\sqrt{2}}{2}t\right)$.

Also, $\overrightarrow{\mathrm{CP}} = \left(t - c(t),\ -\dfrac{\sqrt{2}}{4}t^2 + 4\sqrt{2}\right)$, and since the parabola and circle $C_t$ share a common tangent at P, we have $\vec{u} \cdot \overrightarrow{\mathrm{CP}} = 0$, giving
$$\{t - c(t)\} + \left(-\frac{\sqrt{2}}{2}t\right)\left(-\frac{\sqrt{2}}{4}t^2 + 4\sqrt{2}\right) = 0, \quad t - c(t) + \frac{1}{4}t^3 - 4t = 0$$

From this, $c(t) = \dfrac{1}{4}t^3 - 3t$, and
$$\{r(t)\}^2 = \mathrm{CP}^2 = \{t - c(t)\}^2 + \{f(t)\}^2 = \left(-\frac{1}{4}t^3 + 4t\right)^2 + \frac{1}{8}(t^2 - 16)^2$$
$$= \frac{t^2}{16}(t^2 - 16)^2 + \frac{1}{8}(t^2 - 16)^2 = \frac{1}{16}(t+4)^2(t-4)^2(t^2+2)$$

(2) From (1), circle $C_t$: $\left(x - \dfrac{1}{4}t^3 + 3t\right)^2 + y^2 = \dfrac{1}{16}(t+4)^2(t-4)^2(t^2+2)$, and since $C_t$ passes through the point $(3,\ a)$,
$$\left(3 - \frac{1}{4}t^3 + 3t\right)^2 + a^2 = \frac{1}{16}(t+4)^2(t-4)^2(t^2+2)$$
$$a^2 = \frac{1}{16}(t-16)^2(t^2+2) - \frac{1}{16}(t^3 - 12t - 12)^2 \cdots\cdots\text{\textcircled{1}}$$

Here, let $g(t) = (t^2 - 16)^2(t^2 + 2) - (t^3 - 12t - 12)^2$, then
$$g'(t) = 4t(t^2-16)(t^2+2) + 2t(t^2-16)^2 - 2(t^3-12t-12)(3t^2-12)$$
$$= 2t(t^2-16)\{2(t^2+2)+(t^2-16)\} - 6(t^3-12t-12)(t^2-4)$$
$$= 6t(t^2-16)(t^2-4) - 6(t^3-12t-12)(t^2-4)$$
$$= 6(t^2-4)\{(t^3-16t)-(t^3-12t-12)\} = -24(t+2)(t-2)(t-3)$$

Then, the increase/decrease of $g(t)$ for $0 < t < 4$ is as shown in the table on the right.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$t$ & $0$ & $\cdots$ & $2$ & $\cdots$ & $3$ & $\cdots$ & $4$ \\
\hline
$g'(t)$ & & $-$ & $0$ & $+$ & $0$ & $-$ & \\
\hline
$g(t)$ & $368$ & $\searrow$ & $80$ & $\nearrow$ & $98$ & $\searrow$ & $-16$ \\
\hline
\end{tabular}
\end{center}

Now, the equation \textcircled{1} in $t$ is:
$$16a^2 = g(t) \cdots\cdots\text{\textcircled{2}}$$

Here, since $f(3) = -\dfrac{\sqrt{2}}{4}(9-16) = \dfrac{7}{4}\sqrt{2}$, the real number $a$ takes values $0 < a < \dfrac{7}{4}\sqrt{2}$, so
$$0 < a^2 < \frac{49}{8}, \quad 0 < 16a^2 < 98 \cdots\cdots\text{\textcircled{3}}$$

Therefore, the number of real values of $t$ in the range $0 < t < 4$ satisfying \textcircled{2} is, from \textcircled{3},

\begin{center}
$-6-$ \hfill \copyright\ 電送数学舎 2024
\end{center}



%% Page 13
$0 < 16a^2 < 80 \ (0 < a < \sqrt{5})$ のとき \quad $t$ の個数は 1 個

$16a^2 = 80 \ (a = \sqrt{5})$ のとき \quad $t$ の個数は 2 個

$80 < 16a^2 < 98 \ \left(\sqrt{5} < a < \dfrac{7}{4}\sqrt{2}\right)$ のとき \quad $t$ の個数は 3 個

\bigskip

\noindent\textbf{[Commentary]}

This is a problem combining common tangent lines with applications to differential equations. The content is fundamental, but the calculations are quite involved.



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

Consider the three points $A(1,\ 0,\ 0)$, $B(0,\ 1,\ 0)$, $C(0,\ 0,\ 1)$, and the midpoint $D\!\left(\dfrac{1}{2},\ 0,\ \dfrac{1}{2}\right)$ of segment $AC$. We consider the solid obtained by rotating $\triangle ABD$ once around the $x$-axis.

First, the equation of the plane containing $\triangle ABD$ is $x + y + z = 1$, and

\begin{itemize}
  \item[(a)] The equation of side $AB$ is, for $0 \leq x \leq 1$:
  $$x + y = 1,\quad z = 0 \quad \cdots\cdots \textcircled{1}$$

  \item[(b)] The equation of side $AD$ is, for $\dfrac{1}{2} \leq x \leq 1$: $\quad x + z = 1,\quad y = 0 \quad \cdots\cdots \textcircled{2}$

  \item[(c)] The equation of side $BD$ is, for $0 \leq x \leq \dfrac{1}{2}$: $\quad x = z,\quad 2x + y = 1 \quad \cdots\cdots \textcircled{3}$
\end{itemize}

Now, when $\triangle ABD$ is cut by the plane $x = k$ ($0 \leq k \leq 1$), the cross-section is a line segment. Let $S(k)$ denote the area of the donut-shaped (annular) figure obtained by rotating this line segment around the $x$-axis.

\textbf{(i)} $0 \leq k \leq \dfrac{1}{2}$

The intersection of the plane $x = k$ with side $AB$ is $(k,\ 1-k,\ 0)$ from \textcircled{1}, and the intersection with side $BD$ is $(k,\ 1-2k,\ k)$ from \textcircled{3}, so the cross-section of $\triangle ABD$ is the line segment with these two points as endpoints.

Here, we further consider cases depending on whether the foot of the perpendicular dropped from the point $(k,\ 0,\ 0)$ to the line containing this segment ($x = k$, $y + z = 1 - k$) is $\left(k,\ \dfrac{1-k}{2},\ \dfrac{1-k}{2}\right)$, which lies on the segment or not.

\textbf{(i-i)} $\dfrac{1-k}{2} \leq 1 - 2k \left(0 \leq k \leq \dfrac{1}{3}\right)$

\textit{[Figure: cross-section diagram for case (i-i)]}

In this case, the foot of the perpendicular is not contained in the segment, so the outer radius of the donut is $1 - k$, and the inner radius is $\sqrt{(1-2k)^2 + k^2} = \sqrt{5k^2 - 4k + 1}$, and the area is:
$$S(k) = \pi\{(1-k)^2 - (5k^2 - 4k + 1)\}$$
$$= \pi(-4k^2 + 2k)$$

\textbf{(i-ii)} $\dfrac{1-k}{2} \geq 1 - 2k \left(\dfrac{1}{3} \leq k \leq \dfrac{1}{2}\right)$

\textit{[Figure: cross-section diagram for case (i-ii)]}

In this case, the foot of the perpendicular is contained in the segment, so the outer radius of the donut is $1 - k$, and the inner radius is $\dfrac{1}{\sqrt{2}}(1-k)$, and the area is:
$$S(k) = \pi\left\{(1-k)^2 - \frac{1}{2}(1-k)^2\right\} = \frac{\pi}{2}(1-k)^2$$

\textbf{(ii)} $\dfrac{1}{2} \leq k \leq 1$

The intersection of the plane $x = k$ with side $AB$ is $(k,\ 1-k,\ 0)$ from \textcircled{1}, and the intersection with side $AD$ is $(k,\ 0,\ 1-k)$ from \textcircled{2}, so the cross-section of $\triangle ABD$ is the line segment with these two points as endpoints.



%% Page 15
At this point, the outer radius of the donut shape is $1-k$, the inner radius is $\dfrac{1}{\sqrt{2}}(1-k)$, and its cross-sectional area is,
$$S(k) = \pi\left\{(1-k)^2 - \frac{1}{2}(1-k)^2\right\} = \frac{\pi}{2}(1-k)^2$$

From (i)(ii), $S(k) = \pi(-4k^2 + 2k) \quad \left(0 \leqq k \leqq \dfrac{1}{3}\right)$

$$S(k) = \frac{\pi}{2}(1-k)^2 \quad \left(\frac{1}{3} \leqq k \leqq 1\right)$$

\textit{[Figure: coordinate system with $z$-axis vertical and $y$-axis horizontal, showing a triangular cross-section with labels $1-k$, $\frac{1-k}{2}$, and point at $y=1$, $z=1$]}

From the above, the volume $V$ of the solid is,

$$V = \pi\int_0^{\frac{1}{3}}(-4k^2+2k)\,dk + \frac{\pi}{2}\int_{\frac{1}{3}}^{1}(1-k)^2\,dk$$

$$= \pi\left[-\frac{4}{3}k^3 + k^2\right]_0^{\frac{1}{3}} - \frac{\pi}{6}\left[(1-k)^3\right]_{\frac{1}{3}}^{1} = \frac{5}{81}\pi + \frac{4}{81}\pi = \frac{\pi}{9}$$

\bigskip

\noindent\textbf{[Commentary]}

This is a problem on the volume of a solid of revolution, a frequently appearing type. Despite having a simple diagram and relatively mild calculations, it takes a considerable amount of time nonetheless. This is a problem worth practicing.



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\textbf{6} \hfill (Go to problem page)

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(1) For $f(x) = x^3 + 10x^2 + 20x = x(x^2 + 10x + 20)$, we have $f(n) = n(n^2 + 10n + 20)$.

The integers $n$ for which $f(n)$ is prime are found by letting $p, q, r, s$ be primes:

\begin{itemize}
\item[(i)] $(n,\ n^2 + 10n + 20) = (1,\ p)$ where $f(n) = p$:

$p = 1^2 + 10 \cdot 1 + 20 = 31$, so $f(1) = 31$ is prime.

\item[(ii)] $(n,\ n^2 + 10n + 20) = (-1,\ -q)$ where $f(n) = q$:

$q = -\{(-1)^2 + 10 \cdot (-1) + 20\} = -11$, so $f(-1) = -11$ is not prime.

\item[(iii)] $(n,\ n^2 + 10n + 20) = (r,\ 1)$ where $f(n) = r$:

From $r^2 + 10r + 20 = 1$, we get $r^2 + 10r + 19 = 0$, so no prime $r$ exists.

\item[(iv)] $(n,\ n^2 + 10n + 20) = (-s,\ -1)$ where $f(n) = s$:

From $(-s)^2 + 10(-s) + 20 = -1$, we get $s^2 - 10s + 21 = 0$, so $(s-3)(s-7) = 0$.

Thus $s = 3,\ 7$ are both prime, and $f(-3) = 3$, $f(-7) = 7$.
\end{itemize}

From (i)--(iv), the integers $n$ for which $f(n)$ is prime are $n = 1,\ -3,\ -7$.

\medskip

(2) Let $a, b$ be integer constants. For $g(x) = x^3 + ax^2 + bx = x(x^2 + ax + b)$, we have $g(n) = n(n^2 + an + b)$.

The integers $n$ for which $g(n)$ is prime are found by letting $p, q, r, s$ be primes:

\begin{itemize}
\item[(i)] $(n,\ n^2 + an + b) = (1,\ p)$ where $g(n) = p$:

From $1^2 + a \cdot 1 + b = p$, we get $p = a + b + 1$ \hfill \textcircled{1}

\item[(ii)] $(n,\ n^2 + an + b) = (-1,\ -q)$ where $g(n) = q$:

From $(-1)^2 + a \cdot (-1) + b = -q$, we get $q = -(1 - a + b) = a - b - 1$ \hfill \textcircled{2}

\item[(iii)] $(n,\ n^2 + an + b) = (r,\ 1)$ where $g(n) = r$:

From $r^2 + ar + b = 1$, we get $r^2 + ar + b - 1 = 0$ \hfill \textcircled{3}

\item[(iv)] $(n,\ n^2 + an + b) = (-s,\ -1)$ where $g(n) = s$:

From $(-s)^2 + a(-s) + b = -1$, we get $s^2 - as + b + 1 = 0$ \hfill \textcircled{4}
\end{itemize}

From (i)--(iv), the number of integers $n$ for which $g(n)$ is prime is at most $1 + 1 + 2 + 2 = 6$.

First, consider the case where (iii) and (iv) hold simultaneously. From \textcircled{3} $-$ \textcircled{4}:
$$r^2 - s^2 + a(r + s) - 2 = 0, \quad (r+s)(r - s + a) = 2 \hfill \textcircled{5}$$

Since $r + s \geq 4$, equation \textcircled{5} cannot hold, so (iii) and (iv) cannot hold simultaneously.

Therefore, the number of integers $n$ for which $g(n)$ is prime is at most 4. Below, we consider the case where (i)(ii)(iii) hold simultaneously, and the case where (i)(ii)(iv) hold simultaneously.

\medskip

(a) Case where (i)(ii)(iii) hold simultaneously:

$p = a + b + 1$ \hfill \textcircled{1},\quad $q = a - b - 1$ \hfill \textcircled{2}

Let the two distinct prime solutions $r$ satisfying \textcircled{3} be $r = r_1,\ r_2\ (2 \leq r_1 < r_2)$. Then:
$$r_1 + r_2 = -a \hfill \textcircled{3}', \quad r_1 r_2 = b - 1 \hfill \textcircled{3}''$$

\begin{center}
$-10-$ \hfill \copyright\ 電送数学舎 2024
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\noindent From \textcircled{2}\textcircled{3}$'$ \textcircled{3}$''$,
$$q = -n_1 - n_2 - n_1 n_2 - 1 - 1 = -n_1 n_2 - n_1 - n_2 - 2 < 0 \cdots\cdots\textcircled{6}$$
Since \textcircled{6} does not hold, there is no case where (i)(ii)(iii) hold simultaneously, and there is no case where the number of integers $n$ for which $g(n)$ is prime equals 4.

\noindent (b) The case where (i)(ii)(iv) hold simultaneously
$$p = a + b + 1 \cdots\cdots\textcircled{1}, \quad q = a - b - 1 \cdots\cdots\textcircled{2}$$
Now, let $s = s_1,\ s_2\ (2 \leq s_1 < s_2)$ be distinct primes satisfying \textcircled{4}, then
$$s_1 + s_2 = a \cdots\cdots\textcircled{4}', \quad s_1 s_2 = b + 1 \cdots\cdots\textcircled{4}''$$
From \textcircled{2}\textcircled{4}$'$ \textcircled{4}$''$,
$$q = s_1 + s_2 - s_1 s_2 + 1 - 1 = -s_1 s_2 + s_1 + s_2 = 1 - (s_1 - 1)(s_2 - 1) < 0 \cdots\cdots\textcircled{7}$$
Since \textcircled{7} does not hold, there is no case where (i)(ii)(iv) hold simultaneously, and there is no case where the number of integers $n$ for which $g(n)$ is prime equals 4.

\noindent From (a)(b), the number of integers $n$ for which $g(n)$ is prime is at most 3.

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\noindent\textbf{[Commentary]}

This is a proof problem using prime numbers as the subject. For the number of integers $n$ in part (2), referring to the concrete example in part (1), the argument proceeds in the order: at most 6 $\to$ at most 4 $\to$ at most 3.