# Convex pentagons are similar if conformally equivalent.

The problem:

Suppose two convex pentagons $A$ and $B$ have equal interior angles (that is, $A=A_1A_2A_3A_4A_5$ and $B=B_1B_2B_3B_4B_5$) with $\angle A_j =\angle B_j$ for each $j\in\{1,\ldots,5\}$).

Suppose that $\mbox{int}(A) \approx \mbox{int}(B)$ are conformally equivalent with a biholomorphism $f:\mbox{int}(A) \rightarrow \mbox{int}(B)$ whose continuous extension to the boundary maps $A_i\overset{f}{\mapsto}B_i$.

Show that under these conditions, $A$ and $B$ are similar.

Ideas:

I would suspect the reflection principle would be applicable, but I'm not certain how to work out the proof.

-
Both are conformally equivalent to the open unit disk, and are conformally equivalent to each other in any case. –  Will Jagy May 7 '12 at 21:51
Anyway, that's the Riemann Mapping Theorem for simply connected regions. The thing you are asked to prove is false. I am not entirely sure what would be a sensible question. Where did you get this? –  Will Jagy May 7 '12 at 22:34
I added a crucial hypothesis. The continuous extension to the boundary (which by correspondence of boundaries maps boundary to boundary), should map vertices to vertices. It was an optional exercise suggested after we did a similar problem on annuli in a complex analysis course. –  Marcel T. May 8 '12 at 5:26
Maybe use a Schwarz-Christoffel mapping to identify each pentagon with the upper half plane? Automorphisms of the upper half plane have a specific form. –  WimC May 8 '12 at 5:39