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 $A{A}^{\prime }$ when $A^{\prime }$ ($R\cos a$, $R\sin a$) is made coincident with $A$ by folding the paper. Let $P(x,y)$ be any point on $MN$, then $|P{A}^{\prime }|=|PA|$. 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 $$\sin \theta =\frac{x}{\sqrt{x^2+y^2}},\cos \theta =\frac{y}{\sqrt{x^2+y^2}}.$$ 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 $\frac{(2x-a{)}^2}{R^2}+\frac{4y^2}{R^2-a^2}=1$.