2 (Go to problem page)
For $\triangle ABC$ with area $1$, let $BC = a$, $CA = b$, $AB = c$. Let $h_1$, $h_2$, $h_3$ be the lengths of the perpendiculars dropped from a point X in the plane to lines $BC$, $CA$, $AB$, respectively. Then, $$\triangle ABX = \frac{1}{2}ch_3, \quad \triangle BCX = \frac{1}{2}ah_1, \quad \triangle CAX = \frac{1}{2}bh_2$$
From the condition, $2 \leq \triangle ABX + \triangle BCX + \triangle CAX \leq 3$ $\cdots\cdots$①, so $$2 \leq \frac{1}{2}ch_3 + \frac{1}{2}ah_1 + \frac{1}{2}bh_2 \leq 3, \quad 4 \leq ah_1 + bh_2 + ch_3 \leq 6 \cdots\cdots\text{②}$$
First, when X is in the interior of $\triangle ABC$ or on its sides, $$\triangle ABX + \triangle BCX + \triangle CAX = \triangle ABC = 1$$
From this, ① does not hold, so X lies outside $\triangle ABC$.
Therefore, we divide the exterior of $\triangle ABC$ into 6 regions with boundary lines $AB$, $BC$, $CA$.
First, consider the case where X is on the opposite side of $A$ with respect to line $BC$, on the same side as $B$ with respect to line $CA$, and on the same side as $C$ with respect to line $AB$. Then from $\triangle ABX + \triangle CAX - \triangle BCX = \triangle ABC$,
[Figure: triangle ABC with point X below side BC, showing perpendicular distances $h_1$, $h_2$, $h_3$]$$\frac{1}{2}ch_3 + \frac{1}{2}bh_2 - \frac{1}{2}ah_1 = 1$$ $$ch_3 + bh_2 = 2 + ah_1$$
From ②, $4 \leq 2 + 2ah_1 \leq 6$, giving $\dfrac{1}{a} \leq h_1 \leq \dfrac{2}{a}$ $\cdots\cdots$③
On the other hand, let $h$ be the length of the perpendicular from $A$ to line $BC$. From $\triangle ABC = 1$, we get $\frac{1}{2}ah = 1$, so $h = \dfrac{2}{a}$ $\cdots\cdots$④
[Figure: triangle ABC with shaded region below BC between distances $\frac{1}{a}$ and $\frac{2}{a}$ from BC, with point X in shaded region]From this, the region where X can exist is the shaded region in the figure to the right from ③④, and the area of this region is: $$\left\{\left(\frac{4}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right\} \times \triangle ABC = \frac{7}{4} \cdots\cdots\text{⑤}$$
Next, consider the case where X is on the same side as $A$ with respect to line $BC$, on the opposite side of $B$ with respect to line $CA$, and on the opposite side of $C$ with respect to line $AB$. Then from $\triangle BCX - \triangle ABX - \triangle CAX = \triangle ABC$,
[Figure: triangle ABC with point X above vertex A, showing perpendicular distances $h_1$, $h_2$, $h_3$]$$\frac{1}{2}ah_1 - \frac{1}{2}ch_3 - \frac{1}{2}bh_2 = 1, \quad ch_3 + bh_2 = ah_1 - 2$$
From ②, $4 \leq 2ah_1 - 2 \leq 6$, giving $\dfrac{3}{a} \leq h_1 \leq \dfrac{4}{a}$ $\cdots\cdots$⑥
[Figure: triangle ABC with shaded region above vertex A between distances $\frac{3}{a}$ and $\frac{4}{a}$ from BC, with point X in shaded region]From this, the region where X can exist is the shaded region in the figure to the right from ④⑥, and the area of this region is: $$\left\{1 - \left(\frac{1}{2}\right)^2\right\} \times \triangle ABC = \frac{3}{4} \cdots\cdots\text{⑦}$$
$-2-$ \copyright \ 電送数学舎 2020
%% Page 9 Since the same argument applies to the other 4 regions exterior to $\triangle ABC$, the range in which point $X$ can move is the shaded region in the figure on the right.
Therefore, the area of the region in which $X$ can move is, from \textcircled{5}\textcircled{7}, $$\left(\frac{7}{4}+\frac{3}{4}\right)\times 3 = \frac{15}{2}$$
[Figure: A hexagonal shaded region with an inner triangle $ABC$ and shaded areas extending outward from each side of the triangle, forming a larger hexagonal shape.][Commentary]This is a problem about plane figures, and it is of the type with no guided steps. However, while it is an interesting problem when time is not a concern, it is quite challenging under time constraints.
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\boxed{3
\text{Go to Problem Page}}
(1) For $-1 \leq t \leq 1$, let $x(t) = (1+t)\sqrt{1+t}$ $\cdots\cdots$①, $y(t) = 3(1+t)\sqrt{1-t}$ $\cdots\cdots$②
define the point $\mathrm{P}(x(t),\, y(t))$.
Here, for $-1 < t \leq 1$, from ①②, $$\frac{y(t)}{x(t)} = \frac{3\sqrt{1-t}}{\sqrt{1+t}} = 3\sqrt{\frac{1-t}{1+t}} = 3\sqrt{-1 + \frac{2}{1+t}}$$
From this, $\dfrac{y(t)}{x(t)}$ is strictly decreasing.
(2) Let $f(t)$ be the distance from the origin to $\mathrm{P}$. Then, $$f(t) = \sqrt{\{x(t)\}^2 + \{y(t)\}^2} = \sqrt{(1+t)^2(1+t) + 9(1+t)^2(1-t)}$$ $$= (1+t)\sqrt{(1+t) + 9(1-t)} = \sqrt{2}\,(1+t)\sqrt{5-4t}$$
$$f'(t) = \sqrt{2}\left\{\sqrt{5-4t} - \frac{4(1+t)}{2\sqrt{5-4t}}\right\} = \sqrt{2} \cdot \frac{5-4t-2-2t}{\sqrt{5-4t}} = \frac{3\sqrt{2}(1-2t)}{\sqrt{5-4t}}$$
From this, the increase/decrease of $f(t)$ for $-1 \leq t \leq 1$ is as shown in the table on the right.
| $t$ | $-1$ | $\cdots$ | $\dfrac{1}{2}$ | $\cdots$ | $1$ |
| $f'(t)$ | | $+$ | $0$ | $-$ | |
| $f(t)$ | $0$ | $\nearrow$ | $\dfrac{3}{2}\sqrt{6}$ | $\searrow$ | $2\sqrt{2}$ |
Thus, at $t = \dfrac{1}{2}$, $f(t)$ attains its maximum value $\dfrac{3}{2}\sqrt{6}$.
(3) For $-1 \leq t \leq 1$, from ①②, $$x'(t) = \frac{3}{2}(1+t)^{\frac{1}{2}} = \frac{3}{2}\sqrt{1+t}$$ $$y'(t) = 3\left\{\sqrt{1-t} - \frac{1+t}{2\sqrt{1-t}}\right\}$$ $$= \frac{3}{2} \cdot \frac{2-2t-1-t}{\sqrt{1-t}} = \frac{3(1-3t)}{2\sqrt{1-t}}$$
The increase/decrease of $x(t)$, $y(t)$ is as shown in the table on the right.
| $t$ | $-1$ | $\cdots$ | $\dfrac{1}{3}$ | $\cdots$ | $1$ |
| $x'(t)$ | | $+$ | $0$ | $+$ | |
| $x(t)$ | $0$ | $\nearrow$ | $\dfrac{8}{9}\sqrt{3}$ | $\nearrow$ | $2\sqrt{2}$ |
| $y'(t)$ | | $+$ | $0$ | $-$ | |
| $y(t)$ | $0$ | $\nearrow$ | $\dfrac{4}{3}\sqrt{6}$ | $\searrow$ | $0$ |
Thus, the locus $C$ of $\mathrm{P}$ has the general shape shown by the bold curve in the figure on the right. Note that the point $\mathrm{P}_0$ is the point on $C$ where $f(t)$ found in (2) attains its maximum value $\dfrac{3}{2}\sqrt{6}$.
Then, when the region $D$ enclosed by $C$ and the $x$-axis is rotated $90°$ clockwise about the origin, the swept region becomes the dotted region in the figure on the right, and its area equals the area of region $D$ plus the area of the quarter circle of radius $\mathrm{OP}_0$.
So, letting $S$ be the area of region $D$,
[Figure: Graph showing curve $C$ in the first quadrant with maximum $y$-value $\frac{4}{3}\sqrt{6}$, point $\mathrm{P}_0$ at distance $\frac{3}{2}\sqrt{6}$ from origin, $x$-intercept at $2\sqrt{2}$, and the rotated region extending to $y = -2\sqrt{2}$, with shaded swept area.]$-4-$ \copyright\ 電送数学舎 2020
%% Page 11 $$S = \int_0^{2\sqrt{2}} y\,dx = \int_{-1}^{1} 3(1+t)\sqrt{1-t} \cdot \frac{3}{2}\sqrt{1+t}\,dt$$ $$= \frac{9}{2}\int_{-1}^{1}(1+t)\sqrt{1-t^2}\,dt = \frac{9}{2}\int_{-1}^{1}(\sqrt{1-t^2}+t\sqrt{1-t^2})\,dt$$
Here, noting that $\displaystyle\int_{-1}^{1}\sqrt{1-t^2}\,dt = \frac{1}{2}\cdot 1^2\cdot\pi = \frac{\pi}{2}$, $\quad \displaystyle\int_{-1}^{1}t\sqrt{1-t^2}\,dt = 0$, $$S = \frac{9}{2}\cdot\frac{\pi}{2} = \frac{9}{4}\pi$$
Also, the area of the quarter circle with radius $\mathrm{OP_0}$ is $\dfrac{1}{2}\!\left(\dfrac{3}{2}\sqrt{6}\right)^{\!2}\cdot\dfrac{\pi}{2} = \dfrac{27}{8}\pi$.
Therefore, the area of the region swept through is $\dfrac{9}{4}\pi + \dfrac{27}{8}\pi = \dfrac{45}{8}\pi$.
[Commentary]This is a comprehensive problem on calculus. Since the hints are carefully laid out, the path to the conclusion of part (3) is clear.
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4 Go to problem page(1) Choose $k$ distinct elements from $2^0, 2^1, 2^2, \ldots, 2^{n-1}$ and take their product. Then, let $a_{n,k}$ denote the sum of these products over all ways of choosing $k$ elements. For $n \geq 2$:
$$a_{n,2} = 2^0 \cdot 2^1 + 2^0 \cdot 2^2 + \cdots + 2^0 \cdot 2^{n-1} + 2^1 \cdot 2^2 + \cdots + 2^1 \cdot 2^{n-1} + \cdots + 2^{n-2} \cdot 2^{n-1}$$
$$= \frac{1}{2}\left\{(2^0 + 2^1 + 2^2 + \cdots + 2^{n-1})^2 - (2^0 + 2^2 + 2^4 + \cdots + 2^{2n-2})\right\}$$
$$= \frac{1}{2}\left(\frac{2^n - 1}{2 - 1}\right)^2 - \frac{1}{2} \cdot \frac{2^{2n}-1}{2^2 - 1} = \frac{1}{2}(2^{2n} - 2\cdot 2^n + 1) - \frac{1}{6}(2^{2n} - 1)$$
$$= \frac{1}{3}(2^{2n} - 3 \cdot 2^n + 2)$$
(2) For a natural number $n$, let $f_n(x) = 1 + a_{n,1}x + a_{n,2}x^2 + \cdots + a_{n,n}x^n \cdots\cdots\textcircled{1}$, then
$$f_{n+1}(x) = 1 + a_{n+1,1}x + a_{n+1,2}x^2 + \cdots + a_{n+1,n}x^n + a_{n+1,n+1}x^{n+1} \cdots\cdots\textcircled{2}$$
Now, $a_{n+1,k}$ is the sum of products obtained by choosing $k$ elements from $2^0, 2^1, 2^2, \ldots, 2^{n-1}, 2^n$, summed over all such choices, for $2 \leq k \leq n$:
(i) When $2^n$ is
not chosen:
Choose $k$ elements from $2^0, 2^1, 2^2, \ldots, 2^{n-1}$ and take their products; the sum of products over all such choices can be written as $a_{n,k}$.
(ii) When $2^n$
is chosen:
Choose $k-1$ elements from $2^0, 2^1, 2^2, \ldots, 2^{n-1}$ (other than $2^n$) and take their products; the sum of products over all such choices can be written as $2^n a_{n,k-1}$.
From (i)(ii): $\quad a_{n+1,k} = a_{n,k} + 2^n a_{n,k-1} \quad (k = 2,\ 3,\ \ldots,\ n) \cdots\cdots\textcircled{3}$
Also, for $k=1$ and $k=n+1$: $$a_{n+1,1} = 2^0 + 2^1 + 2^2 + \cdots + 2^{n-1} + 2^n = a_{n,1} + 2^n \cdots\cdots\textcircled{4}$$ $$a_{n+1,n+1} = 2^0 \cdot 2^1 \cdot 2^2 \cdots\cdots 2^{n-1} \cdot 2^n = 2^n a_{n,n} \cdots\cdots\textcircled{5}$$
Substituting \textcircled{3}\textcircled{4}\textcircled{5} into \textcircled{2}:
$$f_{n+1}(x) = 1 + (a_{n,1} + 2^n)x + (a_{n,2} + 2^n a_{n,1})x^2 + (a_{n,3} + 2^n a_{n,2})x^3 + \cdots$$ $$+ (a_{n,n} + 2^n a_{n,n-1})x^n + 2^n a_{n,n} x^{n+1}$$ $$= f_n(x) + 2^n x(1 + a_{n,1}x + a_{n,2}x^2 + \cdots + a_{n,n-1}x^{n-1} + a_{n,n}x^n)$$ $$= f_n(x) + 2^n x\, f_n(x) = (1 + 2^n x)f_n(x) \cdots\cdots\textcircled{6}$$
Therefore, $\dfrac{f_{n+1}(x)}{f_n(x)} = 1 + 2^n x$.
Next, since $f_1(x) = 1 + a_{1,1}x = 1 + 2^0 x$, for $n \geq 2$, from \textcircled{6}:
$$f_n(x) = f_1(x)(1 + 2^1 x)(1 + 2^2 x)\cdots(1 + 2^{n-1}x)$$ $$= (1 + 2^0 x)(1 + 2^1 x)(1 + 2^2 x)\cdots(1 + 2^{n-1}x) \cdots\cdots\textcircled{7}$$
Note that \textcircled{7} also holds for $n = 1$.
%% Page 13 From \textcircled{7}, $f_n(2x) = (1+2^1x)(1+2^2x)(1+2^3x)\cdots(1+2^nx)$, and $$f_{n+1}(x) = (1+2^0x)(1+2^1x)(1+2^2x)\cdots(1+2^{n-1}x)(1+2^nx)$$ $$= (1+2^0x)f_n(2x) = (1+x)f_n(2x) \cdots\cdots\textcircled{8}$$ Therefore, $\dfrac{f_{n+1}(x)}{f_n(2x)} = 1+x$.
(3) From \textcircled{3}, $a_{n+1,k+1} = a_{n,k+1} + 2^n a_{n,k}$ $(k=1,\ 2,\ \cdots,\ n-1)$ $\cdots\cdots\cdots$\textcircled{9}
Also, from \textcircled{1}, $f_n(2x) = 1 + 2a_{n,1}x + 2^2a_{n,2}x^2 + \cdots + 2^n a_{n,n}x^n$, and comparing the coefficients of $x^{k+1}$ on both sides of \textcircled{8}, $$a_{n+1,k+1} = 2^{k+1}a_{n,k+1} + 2^k a_{n,k} \quad (k=1,\ 2,\ \cdots,\ n-1) \cdots\cdots\cdots\textcircled{10}$$
From \textcircled{9}\textcircled{10}, $(2^{k+1}-1)a_{n+1,k+1} = (2^{n+k+1}-2^k)a_{n,k}$, and $$\frac{a_{n+1,k+1}}{a_{n,k}} = \frac{2^{n+k+1}-2^k}{2^{k+1}-1} = \frac{2^k(2^{n+1}-1)}{2^{k+1}-1} \quad (k=1,\ 2,\ \cdots,\ n-1) \cdots\cdots\cdots\textcircled{11}$$
Note that from \textcircled{5}, \textcircled{11} also holds when $k=n$.
[Commentary]This is a challenging problem combining number theory and sequences. In part (2), we considered $a_{n,1},\ a_{n,2},\ a_{n,3},\ \ldots$ concretely, and following the method of deriving the binomial coefficient formula, we constructed a solution that builds a recurrence relation for the coefficients. However, the intended approach of the problem seems to be to set up equation \textcircled{7} directly from the problem statement. The verification work was also quite involved.
%% Page 14
5 Go to problem page(1) Consider the cone $S$ whose base is the circle of radius $1$ centered at the origin in the $xy$-plane, and whose apex is the point $(0,\,0,\,2)$ (including the interior). When $S$ is cut by the plane $z=1$, the cross-section is the interior or boundary of the circle with center $(0,\,0,\,1)$ and radius $\dfrac{1}{2}$.
Now, let $A(1,\,0,\,2)$, and let $\theta$ be any real number. Let $P$ be a point on the base of cone $S$, written as $P(r\cos\theta,\,r\sin\theta,\,0)$ where $0\leq r\leq 1$. Consider the portion $T$ of the cone $S$ swept out by the line segment $AP$, and consider the cross-section of $T$ cut by the plane $z=1$. Let $Q(x,\,y,\,1)$ be the intersection of line segment $AP$ with the plane $z=1$. Since $P$ divides segment $AQ$ externally in the ratio $2:1$, $$(r\cos\theta,\;r\sin\theta,\;0)=(-1+2x,\;2y,\;0)$$ Then the locus of point $Q$ can be expressed as $(-1+2x)^2+(2y)^2=r^2$ and $z=1$, and since $0\leq r\leq 1$ gives $r^2\leq 1$, $$(-1+2x)^2+(2y)^2\leq 1 \quad\text{and}\quad z=1$$ $$\left(x-\frac{1}{2}\right)^2+y^2\leq\frac{1}{4} \quad\text{and}\quad z=1$$
From the above, plotting the cross-sections of $S$ and $T$ by the plane $z=1$ on the plane $z=1$ gives the shaded region shown in the figure on the right (boundary included).
[Figure: Two circles on the $z=1$ plane; the left circle (cross-section of $S$) centered at the origin with radius $\frac{1}{2}$, and the right circle (cross-section of $T$) centered at $(\frac{1}{2},0)$ with radius $\frac{1}{2}$, both shaded.](2) Similarly to (1), let $P(r\cos\theta,\,r\sin\theta,\,k)$ be a point moving on the cross-section of $S$ by the plane $z=k$, and let $Q(x,\,y,\,t)$ be the intersection of line segment $AP$ with the plane $z=t$. Here $0\leq k\leq t\leq 2$.
The radius $r_0$ of the circular cross-section of $S$ by the plane $z=k$ satisfies $(2-k):2=r_0:1$, giving $$r_0=\frac{2-k}{2}$$ so $0\leq r\leq\dfrac{2-k}{2}$.
Since point $P$ divides segment $AQ$ externally in the ratio $(2-k):(t-k)$, $$(r\cos\theta,\;r\sin\theta,\;k)=\left(\frac{(2-k)x-(t-k)}{2-t},\;\frac{(2-k)y}{2-t},\;k\right)$$ Then the locus of point $Q$ can be expressed as $\left\{\dfrac{(2-k)x-(t-k)}{2-t}\right\}^2+\left\{\dfrac{(2-k)y}{2-t}\right\}^2=r^2$ and $z=t$, and since $0\leq r\leq\dfrac{2-k}{2}$ gives $r^2\leq\left(\dfrac{2-k}{2}\right)^2$, $$\left\{\frac{(2-k)x-(t-k)}{2-t}\right\}^2+\left\{\frac{(2-k)y}{2-t}\right\}^2\leq\left(\frac{2-k}{2}\right)^2 \quad\text{and}\quad z=t$$ $$\left(x-\frac{t-k}{2-k}\right)^2+y^2\leq\left(\frac{2-t}{2}\right)^2 \quad\text{and}\quad z=t$$
$-8-$ {\small \copyright\ 電送数学舎\ 2020}
%% Page 15 From this, point Q traces the interior or boundary of a circle on the plane $z = t$ with center $\left(\dfrac{t-k}{2-k},\ 0,\ t\right)$ and radius $\dfrac{2-t}{2}$.
Now, fixing $t$ with $0 \leq t \leq 2$ and letting $k$ vary over $0 \leq k \leq t$, the circle of radius $\dfrac{2-t}{2}$ that is the locus of point Q on $z = t$ has its center moving from $\left(\dfrac{t}{2},\ 0,\ t\right)$ to $(0,\ 0,\ t)$, so the region swept out is the shaded area in the figure on the right. The area of this shaded region is: $$\pi\left(\frac{2-t}{2}\right)^2 + \frac{t}{2}\cdot\frac{2-t}{2}\cdot 2 = \frac{\pi}{4}(2-t)^2 + \frac{1}{2}t(2-t)$$
Therefore, the volume $V$ of the region swept by line segment AP as point P moves over $S$ is: $$V = \int_0^2 \left\{\frac{\pi}{4}(2-t)^2 + \frac{1}{2}t(2-t)\right\}dt = -\frac{\pi}{4}\left[\frac{(2-t)^3}{3}\right]_0^2 + \frac{1}{2}\cdot\frac{1}{6}\cdot 2^3 = \frac{2}{3}\pi + \frac{2}{3}$$
[Figure: On the $z=t$ plane, two overlapping circles; the shaded region shows the swept area with labels $\frac{2-t}{2}$, $\frac{t}{2}$, $-\frac{2-t}{2}$ on the $y$-axis and $1$ on the $x$-axis.][Commentary]This is a frequently appearing type of problem at the University of Tokyo involving the volume of a solid. For part (2), one would naturally use the result from part (1) as a guide; in the solution above, however, we have described it somewhat carefully without omitting overlapping details. Note that the positional relationship among A, P, and Q was handled using external division rather than internal division, in order to reduce the amount of computation.
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6 Go to the problem page(1) For the equation $A\sin 2\theta - \sin(\theta + \alpha) = 0 \cdots\cdots\textcircled{1}$ in $\theta$, let $f(\theta)$ denote the left-hand side of \textcircled{1}: $$f(\theta) = A\sin 2\theta - \sin(\theta + \alpha) \quad (0 \leq \theta < 2\pi)$$
Then, when $A > 1$, $$f\!\left(\frac{\pi}{4}\right) = A - \sin\!\left(\frac{\pi}{4} + \alpha\right) > 0, \quad f\!\left(\frac{3}{4}\pi\right) = -A - \sin\!\left(\frac{3}{4}\pi + \alpha\right) < 0$$ $$f\!\left(\frac{5}{4}\pi\right) = A - \sin\!\left(\frac{5}{4}\pi + \alpha\right) > 0, \quad f\!\left(\frac{7}{4}\pi\right) = -A - \sin\!\left(\frac{7}{4}\pi + \alpha\right) < 0$$
Since $f(\theta)$ is a continuous function, \textcircled{1} has at least one solution in each of the intervals $\dfrac{\pi}{4} < \theta < \dfrac{3}{4}\pi$, $\dfrac{3}{4}\pi < \theta < \dfrac{5}{4}\pi$, $\dfrac{5}{4}\pi < \theta < \dfrac{7}{4}\pi$.
Also, since $f(0) = f(2\pi) = -\sin\alpha$,
(i) When $\sin\alpha \geq 0$:
$f(0) \leq 0$, so \textcircled{1} has at least one solution in the interval $0 \leq \theta < \dfrac{\pi}{4}$.
(ii) When $\sin\alpha < 0$:
$f(2\pi) > 0$, so \textcircled{1} has at least one solution in the interval $\dfrac{7}{4}\pi < \theta < 2\pi$.
From (i)(ii), \textcircled{1} has at least 4 solutions in the range $0 \leq \theta < 2\pi$.
(2) Let a point $\mathrm{Q}$ on the ellipse $C: \dfrac{x^2}{2} + y^2 = 1$ be $\mathrm{Q}(\sqrt{2}\cos\theta,\ \sin\theta)$
[Figure: ellipse $C$ with semi-axes $\sqrt{2}$ and $1$, point $Q$ in first quadrant, point $P$ near origin on $x$-axis]$(0 \leq \theta < 2\pi)$. Then the equation of the tangent to $C$ at $\mathrm{Q}$ is: $$\frac{\sqrt{2}}{2}x\cos\theta + y\sin\theta = 1$$
The normal vector $\overrightarrow{n_1}$ of this tangent is: $$\overrightarrow{n_1} = \left(\frac{\sqrt{2}}{2}\cos\theta,\ \sin\theta\right) = \frac{\sqrt{2}}{2}(\cos\theta,\ \sqrt{2}\sin\theta)$$
Then, the direction vector $\overrightarrow{n_2}$ of the normal line to $C$ at $\mathrm{Q}$ is $\overrightarrow{n_2} = (\sqrt{2}\sin\theta,\ -\cos\theta)$, so the equation of the normal line is $\sqrt{2}\sin\theta(x - \sqrt{2}\cos\theta) - \cos\theta(y - \sin\theta) = 0$, giving: $$\sqrt{2}x\sin\theta - y\cos\theta - \sin\theta\cos\theta = 0, \quad \sin 2\theta - 2\sqrt{2}x\sin\theta + 2y\cos\theta = 0$$
Now, suppose this normal line passes through the point $\mathrm{P}(u,\ v)$ $(2u^2 + v^2 < r^2)$: $$\sin 2\theta - 2\sqrt{2}u\sin\theta + 2v\cos\theta = 0 \cdots\cdots\cdots\textcircled{2}$$
(i) When $(u,\ v) = (0,\ 0)$:
\textcircled{2} becomes $\sin 2\theta = 0$, which has 4 solutions in the range $0 \leq \theta < 2\pi$.
(ii) When $(u,\ v) \neq (0,\ 0)$:
Setting $\cos\alpha = \dfrac{\sqrt{2}u}{\sqrt{2u^2 + v^2}}$, $\sin\alpha = \dfrac{-v}{\sqrt{2u^2 + v^2}}$, \textcircled{2} becomes: $$\sin 2\theta - 2\sqrt{2u^2 + v^2}\,\sin(\theta + \alpha) = 0$$
%% Page 17 $$\frac{1}{2\sqrt{2u^2+v^2}}\sin 2\theta - \sin(\theta+\alpha) = 0 \cdots\cdots\text{③}$$
Here, since $(u,\,v)\neq(0,\,0)$ and $2u^2+v^2 < r^2$ $(0
\frac{1}{2r}$$
Then, for $0 < r \leq \dfrac{1}{2}$, we have $\dfrac{1}{2\sqrt{2u^2+v^2}} > 1$, and in this case from (1), equation ③ has at least 4 solutions in the range $0 \leq \theta < 2\pi$.
From (i)(ii), when $0 < r \leq \dfrac{1}{2}$, for any $(u,\,v)$, equation ② has at least 4 solutions.
Therefore, when $\mathrm{P}(u,\,v)$ is any point inside the region $D: 2x^2+y^2 < r^2$ $\left(0 < r \leq \dfrac{1}{2}\right)$,
there exist at least 4 points Q satisfying the condition, and such $r$ exists.
Now, for $r > \dfrac{1}{2}$, for some $(u,\,v)$ satisfying $2u^2+v^2 = \dfrac{1}{4}$, equation ③ becomes $$\sin 2\theta - \sin(\theta+\alpha) = 0 \cdots\cdots\text{④}$$
Here, considering the case $\alpha = \dfrac{\pi}{4}$, equation ④ becomes $\sin 2\theta - \sin\!\left(\theta+\dfrac{\pi}{4}\right) = 0 \cdots\cdots\text{⑤}$, and $$2\cos\frac{1}{2}\!\left(3\theta+\frac{\pi}{4}\right)\sin\frac{1}{2}\!\left(\theta-\frac{\pi}{4}\right)=0,\quad 2\cos\frac{12\theta+\pi}{8}\sin\frac{4\theta-\pi}{8}=0$$
From $0\leq\theta<2\pi$, we have $\dfrac{\pi}{8}\leq\dfrac{12\theta+\pi}{8}<\dfrac{25}{8}\pi$, $\ -\dfrac{\pi}{8}\leq\dfrac{4\theta-\pi}{8}<\dfrac{7}{8}\pi$, and $$\frac{12\theta+\pi}{8}=\frac{\pi}{2},\ \frac{3}{2}\pi,\ \frac{5}{2}\pi \quad\text{or}\quad \frac{4\theta-\pi}{8}=0$$
Therefore, $\theta = \dfrac{\pi}{4},\ \dfrac{11}{12}\pi,\ \dfrac{19}{12}\pi$, so equation ⑤ has only 3 solutions.
Thus, since the condition is not satisfied when $r > \dfrac{1}{2}$, the maximum value of $r$ is $\dfrac{1}{2}$.
[Commentary]
This is a geometry problem based on an ellipse. Treating (1) as a hint, the policy of parametrically representing point Q in (2) becomes clear. Note that, although omitted in the solution, in the latter half of (1) and (2), one can draw the two graphs $y = A\sin 2\theta$ and $y = \sin(\theta+\alpha)$ to find the intersections.