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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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    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.2012-05-08

1 Answers 1

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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.