Ellipses and Planes

Let the plane be divided into at most $A_n$ parts by $n$ ellipses. The $(n+1)^{th}$ ellipse intersect the $n$ ellipses at most at $4n$ distinct points.Thus the $(n+1)^{th}$ ellipse is divided into $4n$ arcs and every original region is divided into two parts by each arc. So we get,

$A_{n+1}=A_{n}+4n$ and $A_{1}=2$

So after summing it up,we get,

$A_{n}=2n^{2}-2n+2$

For $n=10$ ,we get $A_{10}=182$ .

This was a great problem!!

We will use method of successive differences to solve it. Please bear in mind that this may not work for all such problems, this one is simple, therefore it works here.

Denote maximum number of parts for $n$ ellipses as $a_n$

Difficult part: Find $a_n$ for $n=1,2,3$ and finally, if possible $4$ . I did do this. And got the values as: $2,6,14,26$ respectively.

Now, we can see that the successive differences are as follows:

${ a }_{ 1 }\quad \quad { a }_{ 2 }\quad \quad { a }_{ 3 }\quad \quad { a }_{ 4 }\\ 2\quad \quad \quad 6\quad \quad 14\quad \quad 26\\ \quad \quad 4\quad \quad \quad 8\quad \quad 12\\ \quad \quad \quad \quad 4\quad \quad \quad 4\\ \quad \quad \quad \quad \quad \quad 0$

Since the difference vanishes after two levels, we can conclude that the formula for $a_n$ is a quadratic in $n$ .

Let $a_n=pn^2+qn+r$

Solving for $p,q,r$ using known values of $a_n$ , we get: $p=2,q=-2,r=2$

Hence $a_n=2n^2-2n+2\Rightarrow a_{10}=\boxed{182}$

interesting solution: