Isosceles from regular

First, let's find which regular polygons work.

Any even-sided regular polygon can be converted into one of half as many sides by connecting every other vertex. This means that since a square can be done, every number of sides of the form $2^{n}$ can also be done.

It also means we now only have to consider the odd numbers, since the rest of the evens can be converted to an odd by this process.

An equilateral triangle is isosceles so we get this one for free as well as all numbers of form $3*2^{n}$

The next cool thing is that if you connect any two diagonals that are a power of $2$ away from each other you can convert the rest of this side into isosceles triangles by the same process. The decagon with the problem shows this.

There is only one type of isosceles triangle left to mention: the one on odd-sided polygons that connects two consecutive vertices to the one directly opposite.

This leaves two equal sides, which if they are powers of two will be reducible as well. In other words numbers of the form $2^{m} + 1$ are doable as well as any multiple of this by a power of $2$ .

So the sides of the polygon is any number of the form $(2^{m} + 1)*2^{n}$ for any non-negative integers $n,m$ .

This can also be written as $2^{a}+2^{b}$ .

$2^{13}=8192 < 10000 < 2^{14}=16384$

$8192+2^{10} < 10000 < 8192+2^{11}=\boxed{10240}$