6 Go to the solutions pageAnswer the following questions.
- [(1)] Let $A$, $\alpha$ be real numbers. Consider the equation in $\theta$: $A\sin 2\theta - \sin(\theta + \alpha) = 0$. When $A > 1$, show that this equation has at least 4 solutions in the range $0 \leq \theta < 2\pi$.
- [(2)] Consider the ellipse $C: \dfrac{x^2}{2} + y^2 = 1$ in the coordinate plane. For a real number $r$ satisfying $0 < r < 1$, let $D$ be the region represented by the inequality $2x^2 + y^2 < r^2$. Show that there exists a real number $r$ $(0 < r < 1)$ such that every point $\mathrm{P}$ in $D$ satisfies the following condition. Also, find the maximum value of such $r$.
Condition: There are at least 4 points $\mathrm{Q}$ on $C$ such that the tangent line to $C$ at $\mathrm{Q}$ and the line $\mathrm{PQ}$ are perpendicular to each other.
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1 Go to problem page(1) Consider $ax^2 + bx + c > 0 \cdots\cdots\textcircled{1}$, $bx^2 + cx + a > 0 \cdots\cdots\textcircled{2}$, $cx^2 + ax + b > 0 \cdots\cdots\textcircled{3}$. The solution of this system of inequalities is $x > p$ by assumption. Here, suppose $a < 0$. Then,
$$\lim_{x \to \infty}(ax^2 + bx + c) = \lim_{x \to \infty} x^2\!\left(a + \frac{b}{x} + \frac{c}{x^2}\right) = -\infty$$
From this, for sufficiently large $x$, we have $ax^2 + bx + c < 0$, so the solution of \textcircled{1} cannot contain $x > p$. Therefore, $a \geqq 0$.
Similarly, from \textcircled{2} we get $b \geqq 0$, and from \textcircled{3} we get $c \geqq 0$.
(2) Suppose $a > 0$, $b > 0$, and $c > 0$. Then, in the same way as (1), $$\lim_{x \to -\infty}(ax^2 + bx + c) = \infty, \quad \lim_{x \to -\infty}(bx^2 + cx + a) = \infty, \quad \lim_{x \to -\infty}(cx^2 + ax + b) = \infty$$
From this, for sufficiently small $x$, all of \textcircled{1}\textcircled{2}\textcircled{3} hold. But then the solution of the system of inequalities \textcircled{1}\textcircled{2}\textcircled{3} being $x > p$ is contradicted, so from the conclusion of (1), at least one of $a$, $b$, $c$ is $0$.
(3) From (1)(2), first consider the case $a = 0$, $b \geqq 0$, $c \geqq 0$. From \textcircled{1}\textcircled{2}\textcircled{3},
$$bx + c > 0 \cdots\cdots\cdots\textcircled{4}, \quad bx^2 + cx > 0 \cdots\cdots\cdots\textcircled{5}, \quad cx^2 + b > 0 \cdots\cdots\cdots\textcircled{6}$$
From \textcircled{5}, $x(bx + c) > 0$, so from \textcircled{4}, $x > 0 \cdots\cdots\textcircled{7}$, and
(i) When $b = 0$:
From \textcircled{4}, $c > 0$, so from \textcircled{6}, $cx^2 > 0$, that is, $x \neq 0$. From this, the solution of the system of inequalities \textcircled{4}\textcircled{6}\textcircled{7} is $x > 0$.
(ii) When $b > 0$:
Since $c \geqq 0$, under \textcircled{7}, \textcircled{4} and \textcircled{6} hold, so the solution of the system of inequalities \textcircled{4}\textcircled{6}\textcircled{7} is $x > 0$.
From (i)(ii), the solution of the system of inequalities \textcircled{4}\textcircled{5}\textcircled{6} is $x > 0$.
Also, considering similarly the cases $a \geqq 0$, $b = 0$, $c \geqq 0$ and $a \geqq 0$, $b \geqq 0$, $c = 0$, the solution of the system of inequalities \textcircled{1}\textcircled{2}\textcircled{3} is $x > 0$ in all cases, so $p = 0$.
[Commentary]This is a proof problem on systems of inequalities. Each part is considered by associating graphs.
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