6Consider a coordinate plane with O as the origin. For each non-negative integer $k$, define the vector $\vec{v_k} = \left(\cos\dfrac{2k\pi}{3},\ \sin\dfrac{2k\pi}{3}\right)$. A coin that shows heads or tails each with probability $\dfrac{1}{2}$ is tossed $N$ times, and points $\mathrm{X}_0,\ \mathrm{X}_1,\ \mathrm{X}_2,\ \ldots,\ \mathrm{X}_N$ on the coordinate plane are defined according to the following rules (i), (ii).
- [(i)] $\mathrm{X}_0$ is at O.
- [(ii)] Let $n$ be an integer with $1 \leq n \leq N$. Suppose $\mathrm{X}_{n-1}$ has been determined; then $\mathrm{X}_n$ is defined as follows.
- If the $n$-th coin toss shows heads, $\mathrm{X}_n$ is determined by $\overrightarrow{\mathrm{OX}_n} = \overrightarrow{\mathrm{OX}_{n-1}} + \vec{v_k}$, where $k$ is the number of times tails has appeared from the 1st through the $n$-th coin toss.
- If the $n$-th coin toss shows tails, $\mathrm{X}_n$ is defined to be $\mathrm{X}_{n-1}$.
- [(1)] Let $N = 8$. Find the probability that $\mathrm{X}_8$ is at O.
- [(2)] Let $N = 200$. Let $p_r$ be the probability that $\mathrm{X}_{200}$ is at O and that heads appears exactly $r$ times in the 200 coin tosses in total, where $0 \leq r \leq 200$. Find $p_r$. Also find the value of $r$ for which $p_r$ is maximized.
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1 (Go to problem page)
(1) $f(x) = (\cos x)\log(\cos x) - \cos x + \displaystyle\int_0^x (\cos t)\log(\cos t)\,dt \;\left(0 \leq x < \dfrac{\pi}{2}\right)$ に対して,
$$f'(x) = -(\sin x)\log(\cos x) + (\cos x)\cdot\frac{(-\sin x)}{\cos x} + \sin x + (\cos x)\log(\cos x)$$
$$= -(\sin x - \cos x)\log(\cos x)$$
$$= -\sqrt{2}\sin\!\left(x - \frac{\pi}{4}\right)\cdot\log(\cos x)$$
$f(x)$ is, on $0 \leq x < \dfrac{\pi}{2}$, increasing/decreasing as shown in the table on the right, and has a minimum at $x = \dfrac{\pi}{4}$.
| $x$ | $0$ | $\cdots$ | $\dfrac{\pi}{4}$ | $\cdots$ | $\dfrac{\pi}{2}$ |
| $f'(x)$ | | $-$ | $0$ | $+$ | $\times$ |
| $f(x)$ | $-1$ | $\searrow$ | | $\nearrow$ | $\times$ |
(2) The minimum value is $f\!\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}\log\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} + \displaystyle\int_0^{\frac{\pi}{4}} (\cos t)\log(\cos t)\,dt$, and
$$I = \int_0^{\frac{\pi}{4}} (\cos t)\log(\cos t)\,dt = \Bigl[(\sin t)\log(\cos t)\Bigr]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} (\sin t)\cdot\frac{(-\sin t)}{\cos t}\,dt$$
$$= \frac{1}{\sqrt{2}}\log\frac{1}{\sqrt{2}} + \int_0^{\frac{\pi}{4}} \frac{1-\cos^2 t}{\cos t}\,dt = \frac{1}{\sqrt{2}}\log\frac{1}{\sqrt{2}} + \int_0^{\frac{\pi}{4}} \frac{1}{\cos t}\,dt - \Bigl[\sin t\Bigr]_0^{\frac{\pi}{4}}$$
$$= \frac{1}{\sqrt{2}}\log\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \int_0^{\frac{\pi}{4}} \frac{\cos t}{\cos^2 t}\,dt = \frac{1}{\sqrt{2}}\log\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \int_0^{\frac{\pi}{4}} \frac{\cos t}{1-\sin^2 t}\,dt$$
$$= \frac{1}{\sqrt{2}}\log\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \frac{1}{2}\int_0^{\frac{\pi}{4}} \left(\frac{1}{1-\sin t} + \frac{1}{1+\sin t}\right)\cos t\,dt$$
$$= \frac{1}{\sqrt{2}}\log\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \frac{1}{2}\left[\log\left|\frac{1+\sin t}{1-\sin t}\right|\right]_0^{\frac{\pi}{4}} = \frac{1}{\sqrt{2}}\log\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \frac{1}{2}\log\frac{\sqrt{2}+1}{\sqrt{2}-1}$$
Therefore, $\quad f\!\left(\dfrac{\pi}{4}\right) = \dfrac{2}{\sqrt{2}}\log\dfrac{1}{\sqrt{2}} - \dfrac{2}{\sqrt{2}} + \dfrac{1}{2}\log\dfrac{\sqrt{2}+1}{\sqrt{2}-1} = -\dfrac{\sqrt{2}}{2}\log 2 - \sqrt{2} + \log(\sqrt{2}+1)$
[Commentary]This is a basic problem on integral equations. The calculations are not particularly troublesome.
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2 (Go to problem page)
(1) For the sequence $\{a_n\}$ defined by $a_1 = 1$, $a_{n+1} = a_n^2 + 1$, writing in mod 5: $$a_1 = 1 \equiv 1, \quad a_2 = a_1^2 + 1 \equiv 1^2 + 1 \equiv 2, \quad a_3 = a_2^2 + 1 \equiv 2^2 + 1 \equiv 5 \equiv 0$$ $$a_4 = a_3^2 + 1 \equiv 0^2 + 1 \equiv 1$$
From this, letting $m$ be a non-negative integer, we can predict that $a_{3m+1} \equiv 1$, $a_{3m+2} \equiv 2$, $a_{3m+3} \equiv 0$ $\cdots\cdots$①, so we prove this by mathematical induction.
- [(i)] When $m = 0$: This holds by the computation above.
- [(ii)] When $m = l$: Assuming ① holds, $$a_{3l+4} = a_{3l+3}^2 + 1 \equiv 0^2 + 1 \equiv 1, \quad a_{3l+5} = a_{3l+4}^2 + 1 \equiv 1^2 + 1 \equiv 2$$ $$a_{3l+6} = a_{3l+5}^2 + 1 \equiv 2^2 + 1 \equiv 5 \equiv 0$$ Therefore, it also holds when $m = l+1$.
From (i)(ii), ① holds for all non-negative integers $m$. Therefore, when a positive integer $n$ is a multiple of 3, $a_n$ is a multiple of 5.
(2) Since $a_{n+1} - a_n = a_n^2 - a_n + 1 = \left(a_n - \dfrac{1}{2}\right)^2 + \dfrac{3}{4} > 0$, the sequence $\{a_n\}$ is strictly increasing, and $$1 = a_1 < a_2 < \cdots < a_n < \cdots$$
Now, for some natural number $k$, writing in mod $a_k$: $$a_1 \equiv a_1, \quad a_2 \equiv a_2, \quad \cdots, \quad a_{k-1} \equiv a_{k-1}, \quad a_k \equiv 0$$
Then, for natural number $i$, we can predict that $a_{k+i} \equiv a_i$ $\cdots\cdots$②, so we prove this by mathematical induction.
- [(i)] When $i = 1$: $a_{k+1} = a_k^2 + 1 \equiv 0^2 + 1 \equiv 1 \equiv a_1$
- [(ii)] When $i = j$: Assuming $a_{k+j} \equiv a_j$, $$a_{k+j+1} = a_{k+j}^2 + 1 \equiv a_j^2 + 1 \equiv a_{j+1}$$ Therefore, it also holds when $i = j+1$.
From (i)(ii), ② holds for all natural numbers $i$.
Then, $a_{2k} = a_{k+k} \equiv a_k \equiv 0$, $a_{3k} = a_{2k+k} \equiv a_{2k} \equiv 0$, and so if $n$ is a multiple of $k$, then inductively $a_n \equiv 0$. That is, $a_n$ is a multiple of $a_k$.
Also, when $n$ is not a multiple of $k$, letting the remainder when $n$ is divided by $k$ be $r$ ($1 \leq r \leq k-1$), we get $a_n \equiv a_r$ and $a_n \not\equiv 0$. That is, $a_n$ is not a multiple of $a_k$.
From the above, the necessary and sufficient condition for $a_n$ to be a multiple of $a_k$ is that $n$ is a multiple of $k$.
(3) Since $8091 = 2022 \times 4 + 3$, we have $a_{8091} \equiv a_3 \equiv 5$, $(a_{8091})^2 \equiv 25 \pmod{a_{2022}}$, so $$(a_{8091})^2 = a_{2022} \cdot N + 25 \quad (N \text{ is an integer})$$
From this, the greatest common divisor of $a_{2022}$ and $(a_{8091})^2$ equals the greatest common divisor of $a_{2022}$ and $25$.
Here, since $2022 = 3 \times 674$, first $a_{2022}$ is a multiple of $a_3 = 5$.
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2022 University of Tokyo (Science) First Examination Answer and Explanation
Now, in order to investigate whether $a_{2022}$ is a multiple of 25, writing the following in mod 25: $$a_1 = 1 \equiv 1, \quad a_2 = {a_1}^2 + 1 \equiv 1^2 + 1 \equiv 2, \quad a_3 = {a_2}^2 + 1 \equiv 2^2 + 1 \equiv 5$$ $$a_4 = {a_3}^2 + 1 \equiv 5^2 + 1 = 26 \equiv 1$$
From this, by induction, $a_n \equiv 1$ or $a_n \equiv 2$ or $a_n \equiv 5$, so there is no $n$ for which $a_n \equiv 0$, and therefore $a_{2022}$ is not a multiple of 25.
From the above, the greatest common divisor of $a_{2022}$ and $(a_{8091})^2$ is $5$.
[Commentary]This is a difficult problem involving recurrence relations and integers. Regarding (2), for example $a_4 = 26$, and examining mod $a_4 = \bmod\, 26$, we get $a_1 \equiv 1$, $a_2 \equiv 2$, $a_3 \equiv 5$, $a_4 \equiv 0$, $a_5 \equiv 1$, $a_6 \equiv 2$, $a_7 \equiv 5$, $a_8 \equiv 0$, $\ldots$, a periodicity appears, and we can think based on this.
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3 (Go to problem page)
(1) First, for the square $D: 0 \leq x \leq 3$ and $0 \leq y \leq 3$, the region sufficiently far from all of the points O, A(3, 0), B(3, 3) is the shaded region in the figure on the right. Note that the boundary is included in the region.
In this case, when the point $\mathrm{P}(a,\, a^2)$ on the parabola $y = x^2$ lying in region $D$ is in the shaded region of the figure, the range of values that $a$ can take is $1 \leq a \leq \sqrt{3}$.
[Figure: coordinate plane with square region $0\leq x\leq 3$, $0\leq y\leq 3$, shaded region far from O, A, B, with point P on parabola](2) For a point Q in region $D$ that is sufficiently far from all four points O, A, B, P, the area $f(a)$ of the possible region for Q is:
(a) When $1 \leq a \leq \sqrt{2}$:
The possible region for point Q is the shaded region in the figure on the right, and $$f(a) = 1\cdot(a^2-1)+(a-1)\cdot a^2+(3-a-1)\cdot 1$$ $$+2\cdot(3-a^2-1)$$ $$= a^2-1+a^3-a^2+2-a+4-2a^2$$ $$= a^3-2a^2-a+5$$
[Figure: coordinate plane showing shaded region with labeled coordinates $a-1$, $1$, $a$, $2$, $a+1$ on $x$-axis and $a^2-1$, $a^2$, $a^2+1$ on $y$-axis](b) When $\sqrt{2} \leq a \leq \sqrt{3}$:
The possible region for point Q is the shaded region in the figure on the right, and $$f(a) = 1^2+3\cdot(a^2-1-1)+(a-1)\cdot(3-a^2+1)$$ $$+(3-a-1)\cdot(2-a^2+1)$$ $$= 1+3a^2-6-a^3+a^2+4a-4$$ $$+a^3-2a^2-3a+6$$ $$= 2a^2+a-3$$
[Figure: coordinate plane showing shaded region with labeled coordinates $a-1$, $1$, $a$, $2a+1$, $3$ on $x$-axis and $a^2-1$, $a^2$, $2$ on $y$-axis](3) (a) When $1 \leq a \leq \sqrt{2}$: $\quad f'(a) = 3a^2-4a-1$
Here, the solutions to $f'(a)=0$ are $a = \dfrac{2\pm\sqrt{7}}{3}$, and since $2.6 < \sqrt{7} < 2.7$, $$\frac{2-\sqrt{7}}{3} < 1 < \sqrt{2} < \frac{2+\sqrt{7}}{3}$$
Therefore, $f'(a) < 0$, so $f(a)$ is strictly decreasing.
(b) When $\sqrt{2} \leq a \leq \sqrt{3}$: $\quad f'(a) = 4a+1 > 0$, so $f(a)$ is strictly increasing.
From (a) and (b), since $f(a)$ is continuous at $a = \sqrt{2}$, the minimum occurs at $a = \sqrt{2}$.
[Commentary]This is a region problem that requires carefully reading the problem statement, drawing figures, and also demands attention to detail.
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4 (Go to the problem page)
(1) For the curve $C: y = x^3 - x$ $\cdots\cdots$①, the equation of the line $l$ with slope $m$ passing through the point $\mathrm{P}(a,\, b)$ is: $$l: y - b = m(x - a), \quad y = mx - ma + b \cdots\cdots\text{②}$$
Solving ① and ② simultaneously: $x^3 - x = mx - ma + b$, $\quad x^3 - (m+1)x + ma - b = 0 \cdots\cdots\text{③}$
Now, let $f(x) = x^3 - (m+1)x + ma - b$, then $f'(x) = 3x^2 - (m+1)$.
Here, under the condition $m + 1 > 0$ $(m > -1)$,
setting $k = \sqrt{\dfrac{m+1}{3}}$, the increase/decrease of $f(x)$ is as shown in the table on the right:
| $x$ | $\cdots$ | $-k$ | $\cdots$ | $k$ | $\cdots$ |
| $f'(x)$ | $+$ | $0$ | $-$ | $0$ | $+$ |
| $f(x)$ | $\nearrow$ | | $\searrow$ | | $\nearrow$ |
and thus, $$f(-k) = \frac{2(m+1)}{3}\sqrt{\frac{m+1}{3}} + ma - b, \quad f(k) = -\frac{2(m+1)}{3}\sqrt{\frac{m+1}{3}} + ma - b$$
Since $k^2 = \dfrac{m+1}{3}$, we get $m = 3k^2 - 1$, and thus: $$f(-k) = 2k^3 + 3ak^2 - a - b, \quad f(k) = -2k^3 + 3ak^2 - a - b$$
From this, for sufficiently large $m$, $k$ also becomes sufficiently large, and at that point for any $a$, $b$, we have $f(-k) > 0$ and $f(k) < 0$, so $f(x) = 0$, i.e., ③ has 3 distinct real roots.
In other words, for any point $\mathrm{P}$, there exists a line $l$ through $\mathrm{P}$ that intersects the curve $C$ at 3 distinct points.
(2) Let the three real roots of ③ be $x = \alpha,\, \beta,\, \gamma$ $(\alpha < \beta < \gamma)$. Then: $$\alpha + \beta + \gamma = 0, \quad \alpha\beta + \beta\gamma + \gamma\alpha = -(m+1), \quad -\alpha\beta\gamma = ma - b$$
From this, $\beta = -\alpha - \gamma$, and $-(m+1) = -(\alpha+\gamma)^2 + \gamma\alpha = -\alpha^2 - \alpha\gamma - \gamma^2$, $$m + 1 = \alpha^2 + \alpha\gamma + \gamma^2 \cdots\cdots\text{④}, \quad ma - b = -\alpha\gamma(-\alpha - \gamma) = \alpha\gamma(\alpha + \gamma) \cdots\cdots\text{⑤}$$
Here, from the given conditions: $$\int_{\alpha}^{\beta} \{x^3 - (m+1)x + ma - b\}\,dx = -\int_{\beta}^{\gamma} \{x^3 - (m+1)x + ma - b\}\,dx$$
Therefore, $\displaystyle\int_{\alpha}^{\gamma} \{x^3 - (m+1)x + ma - b\}\,dx = 0$.
Substituting ④⑤: $$\int_{\alpha}^{\gamma} \{x^3 - (\alpha^2 + \alpha\gamma + \gamma^2)x + \alpha\gamma(\alpha+\gamma)\}\,dx = 0$$
$$\left[\frac{1}{4}x^4 - \frac{\alpha^2 + \alpha\gamma + \gamma^2}{2}x^2 + \alpha\gamma(\alpha+\gamma)x\right]_{\alpha}^{\gamma} = 0$$
$$\frac{1}{4}(\gamma^4 - \alpha^4) - \frac{\alpha^2 + \alpha\gamma + \gamma^2}{2}(\gamma^2 - \alpha^2) + \alpha\gamma(\alpha+\gamma)(\gamma - \alpha) = 0$$
Since $\alpha < \gamma$: $$(\gamma^2 + \alpha^2)(\gamma + \alpha) - 2(\alpha^2 + \alpha\gamma + \gamma^2)(\gamma + \alpha) + 4\alpha\gamma(\alpha+\gamma) = 0$$
$-5-$ \copyright\ 電送数学舎 2022
%% Page 12 $$(\alpha+\gamma)\{(\gamma^2+\alpha^2)-2(\alpha^2+\alpha\gamma+\gamma^2)+4\alpha\gamma\}=0,\quad (\alpha+\gamma)(\gamma-\alpha)^2=0$$
Since $\alpha<\gamma$, we have $\alpha+\gamma=0$, and under $\alpha<0<\gamma$, from \textcircled{5}\textcircled{6}, $$m=\alpha^2+\alpha\gamma+\gamma^2-1=\alpha(\alpha+\gamma)+\gamma^2-1=\gamma^2-1,\quad ma-b=0\cdots\cdots\textcircled{6}$$
Therefore, from \textcircled{2}, $l: y=(\gamma^2-1)x$.
At this point, \textcircled{3} gives $x^3-\gamma^2 x=0$, so $x(x-\gamma)(x+\gamma)=0$, and since $\gamma>0$, this has three distinct real solutions $x=0,\,\pm\gamma$.
Then, the range of possible positions of point $\mathrm{P}(a,\,b)$ is, noting from \textcircled{6} that $ma-b=0$ $(m=\gamma^2-1>-1)$, the shaded region in the figure on the right.
[Figure: coordinate plane with shaded region in the second quadrant and lower-right area, bounded by lines through the origin]However, the boundary excluding the origin is not included.
[Commentary]This is a problem about cubic curves that is difficult to write up. Part (1) asks you to prove something that may seem obvious, but showing the signs of the local maximum and local minimum values is the tricky part. Also, in part (2), the lines satisfying the conditions turn out to be lines passing through the origin, giving the expected result, but\ldots.
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5 Go to problem pageLet $\text{P}(s,\, t,\, u)$ ($1 \leq u \leq 2$) be a point on the lateral surface $S$ of the cone obtained by rotating line segment $\text{AB}$ connecting points $\text{A}(0,\, 0,\, 2)$ and $\text{B}(1,\, 0,\, 1)$ once around the $z$-axis. Since $\angle \text{BAO} = \dfrac{\pi}{4}$,
$$\overrightarrow{\text{AP}} \cdot \overrightarrow{\text{AO}} = |\overrightarrow{\text{AP}}||\overrightarrow{\text{AO}}|\cos\frac{\pi}{4}$$
Here, since $\overrightarrow{\text{AP}} = (s,\, t,\, u-2)$, $\overrightarrow{\text{AO}} = (0,\, 0,\, -2)$,
$$-2(u-2) = \sqrt{s^2 + t^2 + (u-2)^2} \cdot 2 \cdot \frac{1}{\sqrt{2}}$$
Since $1 \leq u \leq 2$, squaring both sides gives $2(2-u)^2 = s^2 + t^2 + (u-2)^2$, and thus
$$s^2 + t^2 = (2-u)^2 \quad \cdots\cdots \textcircled{1}$$
Now, fixing point $\text{P}$, point $\text{Q}$ on the $xy$-plane moves while satisfying $\text{PQ} = 2$, so point $\text{Q}$ traces a circle on the base of the cone with vertex $\text{P}$ and slant height $2$.
Furthermore, the midpoint $\text{M}(x,\, y,\, z)$ of segment $\text{PQ}$ satisfies $\text{PM} = 1$, so on the plane $z = \dfrac{u}{2}$, it traces a circle with center $\left(s,\, t,\, \dfrac{u}{2}\right)$ and radius $\sqrt{1^2 - \left(\dfrac{u}{2}\right)^2} = \dfrac{1}{2}\sqrt{4 - u^2}$, giving
$$(x - s)^2 + (y - t)^2 = \frac{1}{4}(4 - u^2) \quad \cdots\cdots \textcircled{2}$$
Then, the condition for $s$, $t$ satisfying \textcircled{1} and \textcircled{2} to exist on the plane $z = \dfrac{u}{2}$ can be found as the condition for the two circles \textcircled{1} and \textcircled{2} in the $st$-plane to have a common point.
Circle \textcircled{1} has center $(0,\, 0)$ and radius $2 - u$, circle \textcircled{2} has center $(x,\, y)$ and radius $\dfrac{1}{2}\sqrt{4 - u^2}$, and since the distance between the centers of the two circles is $\sqrt{x^2 + y^2}$,
$$\left|(2-u) - \frac{1}{2}\sqrt{4-u^2}\right| \leq \sqrt{x^2+y^2} \leq (2-u) + \frac{1}{2}\sqrt{4-u^2}$$
$$\left\{(2-u) - \frac{1}{2}\sqrt{4-u^2}\right\}^2 \leq x^2 + y^2 \leq \left\{(2-u) + \frac{1}{2}\sqrt{4-u^2}\right\}^2 \quad \cdots\cdots \textcircled{3}$$
From this, the cross-section of the region $K$ through which midpoint $\text{M}$ can pass, cut by the plane $z = \dfrac{u}{2}$, forms a donut (annular) shape from \textcircled{3}. Setting its area as $S(z)$,
$$S(z) = \pi\left\{(2-u) + \frac{1}{2}\sqrt{4-u^2}\right\}^2 - \pi\left\{(2-u) - \frac{1}{2}\sqrt{4-u^2}\right\}^2$$
$$= 2\pi(2-u)\sqrt{4-u^2}$$
Therefore, the volume $V$ of $K$, since $z = \dfrac{u}{2}$,
$$V = \int_{\frac{1}{2}}^{1} S(z)\,dz = 2\pi\int_{1}^{2}(2-u)\sqrt{4-u^2}\cdot\frac{1}{2}\,du = \pi\int_{1}^{2}(2-u)\sqrt{4-u^2}\,du$$
$$= 2\pi\int_{1}^{2}\sqrt{4-u^2}\,du - \pi\int_{1}^{2}u\sqrt{4-u^2}\,du$$
$-7-$ \copyright\ 電送数学舎 2022
%% Page 14 Here, $\displaystyle\int_1^2 \sqrt{4-u^2}\, du = \frac{1}{2}\cdot 2^2 \cdot \frac{\pi}{3} - \frac{1}{2}\cdot 1\cdot \sqrt{3} = \frac{2}{3}\pi - \frac{\sqrt{3}}{2}$
$$\int_1^2 u\sqrt{4-u^2}\, du = \left[-\frac{1}{3}(4-u^2)^{\frac{3}{2}}\right]_1^2 = -\frac{1}{3}\cdot(-3\sqrt{3}) = \sqrt{3}$$
Therefore, $V = 2\!\left(\dfrac{2}{3}\pi - \dfrac{\sqrt{3}}{2}\right)\pi - \sqrt{3}\pi = \dfrac{4}{3}\pi^2 - 2\sqrt{3}\pi$.
[Commentary]This is a frequently appearing problem asking for the volume of a swept region. The solution above solves it using algebraic manipulation.
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\boxed{6
\text{Go to Problem Page}}
For non-negative integers $k$, we have $\vec{v_k} = \left(\cos\dfrac{2k\pi}{3},\ \sin\dfrac{2k\pi}{3}\right)$. Let $l$ be a non-negative integer, then: $$\vec{v_{3l}} = (1,\ 0), \quad \vec{v_{3l+1}} = \left(-\frac{1}{2},\ \frac{\sqrt{3}}{2}\right), \quad \vec{v_{3l+2}} = \left(-\frac{1}{2},\ -\frac{\sqrt{3}}{2}\right)$$
From this, $\vec{v_{3l}} + \vec{v_{3l+1}} + \vec{v_{3l+2}} = \vec{0}$.
Now, toss a coin $N$ times and define points $X_0 = O,\ X_1,\ X_2,\ \ldots,\ X_N$ on the coordinate plane by the following rule.
First, when heads appears: let $a$ be the number of moves by $\vec{v_{3l}}$, $b$ be the number of moves by $\vec{v_{3l+1}}$, and $c$ be the number of moves by $\vec{v_{3l+2}}$. Also, when tails appears, no movement occurs, so: $$\overrightarrow{OX_N} = a\vec{v_{3l}} + b\vec{v_{3l+1}} + c\vec{v_{3l+2}} = \left(a - \frac{1}{2}b - \frac{1}{2}c,\ \frac{\sqrt{3}}{2}b - \frac{\sqrt{3}}{2}c\right)$$
(1) When $N = 8$, $\overrightarrow{OX_8} = \vec{0}$ requires $a\vec{v_{3l}} + b\vec{v_{3l+1}} + c\vec{v_{3l+2}} = \vec{0}$, which gives: $$a = b = c$$ Then, from $0 \leq a + b + c \leq 8$, we get $a = b = c = 0$,\ $a = b = c = 1$,\ $a = b = c = 2$.
Here, denoting heads by $\bigcirc$ and tails by $\times$ in the coin tosses:
$\bullet$ When $a = b = c = 0$:
$\times$ appears 8 times, so only the case ``$\times\times\times\times\times\times\times\times$'' applies, and its probability is $\dfrac{1}{2^8}$.
$\bullet$ When $a = b = c = 1$:
$\times$ appears 5 times. For $\vec{v_i}\ (0 \leq i \leq 5)$: the move of $\vec{v_{3l}}$ ($\vec{v_0}$ or $\vec{v_3}$) occurs 1 time, the move of $\vec{v_{3l+1}}$ ($\vec{v_1}$ or $\vec{v_4}$) occurs 1 time, and the move of $\vec{v_{3l+2}}$ ($\vec{v_2}$ or $\vec{v_5}$) occurs 1 time.
[Figure: diagram showing sequence of moves with $\times$ and $\bigcirc$ symbols for $v_0$ through $v_5$]Then there are $2^3 = 8$ cases, and the probability is $\dfrac{8}{2^8}$.
$\bullet$ When $a = b = c = 2$:
$\times$ appears 2 times. For $\vec{v_i}\ (0 \leq i \leq 2)$: the move of $\vec{v_{3l}}$ ($\vec{v_0}$) occurs 2 times, the move of $\vec{v_{3l+1}}$ ($\vec{v_1}$) occurs 2 times, and the move of $\vec{v_{3l+2}}$ ($\vec{v_2}$) occurs 2 times. Then only the case ``$\bigcirc\bigcirc\times\bigcirc\bigcirc\times\bigcirc\bigcirc$'' applies, and its probability is $\dfrac{1}{2^8}$.
Therefore, the probability that $X_8 = O$ is: $$\frac{1}{2^8} + \frac{8}{2^8} + \frac{1}{2^8} = \frac{10}{2^8} = \frac{5}{128}$$
(2) When $N = 200$, $\overrightarrow{OX_{200}} = \vec{0}$ requires $a = b = c$. Let $p_r$ be the probability that heads appears exactly $r$ times, where $0 \leq r \leq 200$.
First, since the number of heads $r = a + b + c = 3a$, when $r$ is not a multiple of 3, $p_r = 0$.
Next, when $r$ is a multiple of 3, let $0 \leq k \leq 66$ and set $r = 3k$, then:
$-9-$ \copyright\ 電送数学舎 2022
%% Page 16 $$a = b = c = k \quad (0 \leqq k \leqq 66)$$
Thus, $\times$ occurs $200 - 3k$ times, and for $\vec{v_i}$ $(0 \leqq i \leqq 200 - 3k)$: the moves $\overrightarrow{v_{3l}}$ $(\vec{v_0},\ \vec{v_3},\ \cdots,\ \overrightarrow{v_{198-3k}})$ occur $k$ times, the moves $\overrightarrow{v_{3l+1}}$ $(\vec{v_1},\ \vec{v_4},\ \cdots,\ \overrightarrow{v_{199-3k}})$ occur $k$ times, and the moves $\overrightarrow{v_{3l+2}}$ $(\vec{v_2},\ \vec{v_5},\ \cdots,\ \overrightarrow{v_{200-3k}})$ occur $k$ times.
Here, letting the number of moves $\vec{v_0},\ \vec{v_3},\ \cdots,\ \overrightarrow{v_{198-3k}}$ be $a_0,\ a_1,\ \cdots,\ a_{66-k}$ respectively, $$a_0 + a_1 + \cdots + a_{66-k} = k \quad (a_0 \geqq 0,\ a_1 \geqq 0,\ \cdots,\ a_{66-k} \geqq 0)$$ Then the number of combinations $(a_0,\ a_1,\ \cdots,\ a_{66-k})$ is ${}_{67-k}\mathrm{H}_k = {}_{66}\mathrm{C}_k$ (ways).
Similarly, letting the number of moves $\vec{v_1},\ \vec{v_4},\ \cdots,\ \overrightarrow{v_{199-3k}}$ be $b_0,\ b_1,\ \cdots,\ b_{66-k}$ respectively, $$b_0 + b_1 + \cdots + b_{66-k} = k \quad (b_0 \geqq 0,\ b_1 \geqq 0,\ \cdots,\ b_{66-k} \geqq 0)$$ Then the number of combinations $(b_0,\ b_1,\ \cdots,\ b_{66-k})$ is ${}_{67-k}\mathrm{H}_k = {}_{66}\mathrm{C}_k$ (ways).
Furthermore, letting the number of moves $\vec{v_2},\ \vec{v_5},\ \cdots,\ \overrightarrow{v_{200-3k}}$ be $c_0,\ c_1,\ \cdots,\ c_{66-k}$ respectively, $$c_0 + c_1 + \cdots + c_{66-k} = k \quad (c_0 \geqq 0,\ c_1 \geqq 0,\ \cdots,\ c_{66-k} \geqq 0)$$ Then the number of combinations $(c_0,\ c_1,\ \cdots,\ c_{66-k})$ is ${}_{67-k}\mathrm{H}_k = {}_{66}\mathrm{C}_k$ (ways).
Therefore, when $r$ is a multiple of 3, the probability $p_r$ that heads appears $r = 3k$ times is, $$p_r = \frac{({}_{66}\mathrm{C}_k)^3}{2^{200}} = \frac{\left({}_{66}\mathrm{C}_{\frac{r}{3}}\right)^3}{2^{200}}$$
At this point, $\dfrac{p_{3(k+1)}}{p_{3k}} = \dfrac{({}_{66}\mathrm{C}_{k+1})^3}{({}_{66}\mathrm{C}_k)^3} = \left(\dfrac{{}_{66}\mathrm{C}_{k+1}}{{}_{66}\mathrm{C}_k}\right)^3$, and $$\frac{{}_{66}\mathrm{C}_{k+1}}{{}_{66}\mathrm{C}_k} - 1 = \frac{66!}{(k+1)!\,(65-k)!} \cdot \frac{k!\,(66-k)!}{66!} - 1 = \frac{66-k}{k+1} - 1 = \frac{65-2k}{k+1}$$
When $k < \dfrac{65}{2}$ $(k \leqq 32)$, $\dfrac{{}_{66}\mathrm{C}_{k+1}}{{}_{66}\mathrm{C}_k} > 1$; when $k > \dfrac{65}{2}$ $(k \geqq 33)$, $\dfrac{{}_{66}\mathrm{C}_{k+1}}{{}_{66}\mathrm{C}_k} < 1$, so $$p_0 < p_3 < p_6 < \cdots < p_{96} < p_{99} > p_{102} > \cdots > p_{198}$$ From the above, $p_r$ is maximized when $r = 99$.
[Commentary]This is a difficult probability problem that requires careful reading comprehension. The key point is to use the relation $\overrightarrow{v_{3l}} + \overrightarrow{v_{3l+1}} + \overrightarrow{v_{3l+2}} = \vec{0}$ as a basis, investigate the specific patterns of $\bigcirc$ and $\times$, and generalize.