# 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.

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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

Let $f_1,f_2$ be conformal maps of the upper half-plane onto your polygons, and $f$ is your map between the polygons. Then $g:=f_1^{-1}\circ f\circ f_2$ is a conformal automorphism of the upper half-plane. Now consider the functions $f_2$ and $f_1\circ g$. Both map the upper half-plane to polygons, and there are $5$ points $a_1,\ldots,a_5$ on the real line such that for each $k$, both $f_2$ and $f_1\circ g$ map $a_k$ to vertices of their respective polygons, and the interior angles at these two vertices are the same angle.
Now both $f_2$ and $f_1\circ g$ must be represented by the Schwarz-Christoffel formula with the same singularities and same angles. But angles and and singularities determine the Schwarz--Christoffel formula up to a composition with an affine map. Therefore $$f_2=Af_1\circ g+B$$ which proves the statement.
Remarks. 5 is irrelevant. Convexity is also irrelevant. All we need is that the conformal map $f$ between the polygons sends vertices to vertices and the interior angles at the corresponding vertices are equal.