I have been trying to prove Cauchy's integral formula "my way" and it didn't go well, but I found out another interesting geometry problem. If one could prove that my hypotheses are right, I would be able to finish my proof, which would be way more complicated than I initially thought.
The problem can be explained as maintaining a set of convex sets of euclidean plane that cover the given circle. Under "circle", I assume a closed circle defined by its center $c$ and radius $r$, the set: $D(c,r)=\{x\in\mathbb{E}^2\mid|x-c|\leqslant r\}$.
The problem:
Is there a positive real constant $K>1$ and a positive real constant $W>1$ such that for any given circle $B$ and any family of circles $\{A_i=D(c_i,r_i)\mid i\in I\}$ that:
cover the circle $B$: $\bigcup\{A_i\mid i\in I\}\supseteq B$ and
for every $i\in I$, $c_i\in B$,
there exists another family of sets $\{P_j\mid j\in J\}$ such that:
For every $j\in J$, the set $P_j$ is compact and convex,
For every two different indices $j_1,j_2\in J$ holds that $\emptyset=\text{int}P_{j_1}\cap\text{int}P_{j_2}$,
They are an exact cover of $B$: $\bigcup\{P_j\mid j\in J\}=B$,
For every $j\in J$ there exists $i\in I$ so that $c_i\in P_j\subseteq D(c_i,K\cdot r_i)$,
- (this is an extra property) For every $j\in J$, the maximal width $M$ and the minimal width $m$ of $P_j$ form a ratio smaller than $W$: $\frac{M}{m}\leqslant W$?
The fifth condition is additional. If you can prove the theorem for only first four conditions, it is still good. The width of a figure $F$ here represents the minimal width of a strip (the closure of space between two parallel lines) of fixed direction that can completely cover the figure $F$. The minimal width and the maximal width here are taken as extremes when the direction is the parameter.