China Mathematical Competition (Shaanxi)

Establish an $xy$-coordinate system as in the diagram with $A(a, 0)$ given. Then the crease line $MN$ is the perpendicular bisector of segment $AA'$ when $A'$ ($R\cos a$, $R\sin a$) is made coincident with $A$ by folding the paper. Let $P(x, y)$ be any point on $MN$, then $|PA'| = |PA|$. That is

$$(x - R\cos a)^2 + (y - R\sin a)^2 = (x - a)^2 + y^2.$$

Then
$$\frac{x\cos x + y\sin a}{\sqrt{x^2 + y^2}} = \frac{R^2 - a^2 + 2ax}{2R\sqrt{x^2 + y^2}}.$$

We get
$$\sin(\theta + a) = \frac{R^2 - a^2 + 2ax}{2R\sqrt{x^2 + y^2}},$$
where
$$\sin \theta = \frac{x}{\sqrt{x^2 + y^2}}, \quad \cos \theta = \frac{y}{\sqrt{x^2 + y^2}}.$$
So
$$\left| \frac{R^2 - a^2 + 2ax}{2R\sqrt{x^2 + y^2}} \right| \le 1.$$
Squaring both sides, we get
$$\frac{(2x-a)^2}{R^2} + \frac{4y^2}{R^2 - a^2} \ge 1.$$
So the set we want consists of all of the points on the border of or outside the ellipse $\frac{(2x-a)^2}{R^2} + \frac{4y^2}{R^2 - a^2} = 1$.