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A constructible polygon is a polygon whose vertices can be obtained by a compass-and-straightedge construction from the specified initial data. The classical question for a regular polygon asks which numbers of sides permit an exact construction from a given side or circumscribed circle.

Build a closed annulus from two nested $n$-sided polygons, $n\ge 3$, with $n$ radial edges joining corresponding vertices. There are $2n$ vertices, $3n$ edges and $n$ quadrilateral topological cells, giving $\chi =2n-3n+n=0$. The central missing region and the exterior are not faces of the annulus. A deformation retraction onto a circle gives $(b_0,b_1,b_2)=(1,1,0)$, agreeing with the cell count. This is a useful reduction: the annulus is two-dimensional, yet its homology is that of a one-dimensional cycle graph.

A corner of a polygon or polyhedron. In a polygon, it is the shared endpoint of consecutive polygon edges.

A left group action of a group $G$ on a set $X$ is a function $(g,x)\mapsto g\cdot x$ satisfyingwhere $e$ is the identity element of $G$. The second rule means that $h$ moves the point first and $g$ moves it next. These rules let group multiplication describe consistent rearrangements of a set; see Ackerman's algebra notes, definition 0.4.1.1.

For each $g$, the function $x\mapsto g\cdot x$ is a bijection: its inverse is $x\mapsto g^{-1}\cdot x$, using the inverse element of $g$. Equivalently, a group action is a group homomorphism $\rho :G\to \mathrm{Sym}(X)$ into the symmetric group of all permutations of $X$, with $\rho (g)(x)=g\cdot x$ and $\rho (gh)=\rho (g)\circ \rho (h)$. See MIT's group-action notes, definition 23.1.

For a concrete group action, put the vertices of a polygon of a square at $A=(1,1)$, $B=(-1,1)$, $C=(-1,-1)$ and $D=(1,-1)$. Let $r(x,y)=(-y,x)$ be a counterclockwise quarter-turn rotation and let $s(x,y)=(y,x)$ be a reflection. Their eight permutations of the vertices of a polygon form the dihedral groupHere $D_4$ means the symmetry group of a four-sided polygon, with group order eight; the alternative convention calls this same group $D_8$. Products apply the rightmost operation first. For example, $s\cdot B=D$ and $(rs)\cdot B=r\cdot D=A$. The same group also has separate group actions on the sides of a polygon, the diagonals of a polygon, and the singleton set containing the centroid of the square. Specifying $X$ matters: changing what is moved changes the orbits of a group action and the stabilizer subgroups.

A group action need not be a faithful group action. For example, the cyclic group $C_4=\mathbb{Z}/4\mathbb{Z}$ acts on $X=\{0,1\}$ by $k\cdot x=x+k(\mathrm{mod}2)$, using modular arithmetic. Only the parity of $k$ matters: $0,2$ fix both points and $1,3$ exchange them. The associated group homomorphism $C_4\to \mathrm{Sym}(X)$ has group kernel $\{0,2\}$, of group order two. Either point has a group orbit of size two and a stabilizer subgroup of group order two, so the orbit-stabilizer theorem gives $4=2\cdot 2$.

The useful questions are which positions can be exchanged, and which transformations survive after imposing a constraint such as keeping one marked point fixed. An orbit of a group action answers the first question; a stabilizer subgroup answers the second. This turns a list of transformations into a count of distinct configurations, and explains why several transformations can produce the same configuration.

A polygon with six sides and six vertices. A regular hexagon has equal sides and angles, with each interior angle equal to $2\pi /3$. Its symmetry group consists of six rotations and six reflections, forming a dihedral group of order twelve.

Put ${\alpha }_j=1-{\beta }_j$, so the interior angle at the $j$th polygon vertex is $\pi {\alpha }_j$. A conformal map straightens that corner by a power: locally,This follows by composing with the power $1/{\alpha }_j$ that maps the corner to a half-plane and using Schwarz reflection along its straight edges. The reflected map has a nonzero first derivative in the straightened coordinates.

For an upper-half-plane branch of the complex logarithm, defineAlong each open interval of the real axis, $P$ has constant argument. Crossing $x_j$ from left to right decreases $\arg (z-x_j)$ from $\pi$ to zero, so the argument of $P$ increases by ${\beta }_j\pi$. This is exactly the exterior turning angle of the polygon boundary. Thus $f^{\prime }/P$ has one constant boundary argument on all the intervals. The corner asymptotics remove its apparent singularity at each prevertex.

For a bounded polygon, the sum of exterior angles gives ${\sum }_j{\beta }_j=2$. Since every vertex has a finite prevertex, infinity maps to an ordinary point of an edge. In the local coordinate $1/z$, Schwarz reflection gives $f(z)=f(\infty )+d/z+O(z^{-2})$ with $d\ne 0$. Hence $f^{\prime }/P$ has a finite nonzero limit at infinity. After multiplication by one unit complex number, its boundary values are real. Reflecting across the real axis extends it to an entire function, including each removed prevertex, and it is bounded at infinity. Liouville's theorem makes it constant. Therefore the Schwarz-Christoffel formula isIntegration introduces one further additive constant.

The Möbius transformations preserving the upper half-plane have three real parameters:with common nonzero scaling irrelevant. They can send any cyclically ordered triple of boundary points to another such triple. Consequently three prevertices may be fixed independently; the other $n-3$ are constrained by the polygon's shape, through its side-length ratios. The constants of integration and multiplication account for translation, rotation and scale of the image, rather than further prevertex freedom.

If the last prevertex is infinity, use $1/z$ as its local coordinate. Its factor is absorbed into the overall normalization, leavingFor a finite image vertex at infinity's preimage, this has asymptotic order $z^{{\beta }_n-2}$, because $f(z)-f(\infty )$ is of order $z^{-(1-{\beta }_n)}$. Thus the omitted factor is not evaluated by substituting infinity into the finite product; its angle is encoded by the behavior at infinity.

For solid modelling, a polyhedron is a solid whose boundary consists of finitely many planar polygons, called faces, meeting along straight edges and at vertices of a polyhedron. A face is not merely a list of arbitrary spatial points: its boundary must lie in one plane and bound a valid planar region. The boundary is closed and consistently oriented so that inside and outside are defined.

For example, a tetrahedron has four triangular faces, six edges and four vertices of a polyhedron. A more general polyhedron need not be convex and can have handles or cavities. The concise definition is

Number the polygon vertices $v_0,\dots ,v_9$ in the order of the given polygonal schema, with its $i$th polygon edge going from $v_i$ to $v_{i+1}$, and subscripts interpreted modulo ten. Equal labels identify equal arrow endpoints. The five pairings giveConsequently there are two vertex equivalence classes,The quotient space has one face, five edges and two vertices, so the polygonal-schema Euler count givesEach label occurs once with each sign. Hence every pairing reverses the direction induced by traversing the boundary of the polygon, so an orientation of a surface on its interior extends consistently across every glued edge. Thus $X$ is orientable. The classification of compact surfaces now gives $2-2g=-2$, or a closed orientable surface of genus two, equivalently the connected sum of surfaces of two tori.

Take a real modulus $0<k<1$ and complementary modulus $k^{\prime }=\sqrt{1-k^2}$. The Jacobi elliptic sine is constructed by inverting a Schwarz-Christoffel mapping. The main reason for the Schwarz-Christoffel derivative is local angle behavior: if a prevertex $a_j$ corresponds to a polygon angle $\pi {\alpha }_j$, straightening that corner gives $U(w)-U(a_j)\sim C_j(w-a_j{)}^{{\alpha }_j}$. Hence $U^{\prime }$ has exponent ${\alpha }_j-1$. Along each straight boundary side its argument is constant. Dividing $U^{\prime }$ by ${\prod }_j(w-a_j{)}^{{\alpha }_j-1}$ removes those angle changes; Schwarz reflection extends the quotient across the real boundary and the corner singularities. With all vertices finite, its behavior at infinity is also regular, because the sum of the derivative exponents is $-2$. It is therefore a nonzero constant on the sphere. This yieldsThis argument explains the exponents, the branch choices and the role of boundary reflection, rather than just writing down the formula.

For a rectangle, four angles are $\pi /2$, so choose prevertices $-1/k,-1,1,1/k$ and normalize the inverse map asUse the branch positive on $(-1,1)$. SetThe two middle corner values are $U(\pm 1)=\pm K$. The right side length isas the substitution $t=(1-k^{\prime 2}{s}^2{)}^{-1/2}$ shows. The selected branch makes the increment along that side $i{K}^{\prime }$. Thus the other corner values are $K+i{K}^{\prime }$ and $-K+i{K}^{\prime }$. The two boundary intervals through infinity have combined horizontal length $2K$; the substitution $t=1/(ks)$ gives length $K$ from $1/k$ to infinity. In particular $U(\infty )=i{K}^{\prime }$. As in the square mapping, the boundary goes once around a rectangle, and the argument principle gives a conformal bijectionIts inverse $s(z)=\mathrm{sn}(z,k)$ maps that rectangle onto the complex upper half-plane, and near zero has $s(0)=0$, $s^{\prime }(0)=1$. Differentiating the inverse relation givesThese identities continue meromorphically.

All four rectangle edges have real images, so Schwarz reflection extends $s$ across them. Reflection in the bottom edge gives $s(\bar{z})=\overline{s(z)}$. Reflection in the right edge, combined with the first reflection, gives $s(2K-z)=s(z)$. The even inverse-integral derivative gives $s(-z)=-s(z)$, and henceReflection in the top edge and then the bottom gives $s(z+2i{K}^{\prime })=s(z)$. The resulting period lattice of the Jacobi elliptic sine isTo see the poles, use $U^{\prime }(w)\sim -1/(k{w}^2)$ near the boundary point infinity. Then $U(w)-i{K}^{\prime }\sim 1/(kw)$, so the inverse has a simple pole at $i{K}^{\prime }$ with residue $1/k$. The real half-period shift gives another pole at $2K+i{K}^{\prime }$ with opposite residue. Reflection tiles the plane with copies of the rectangle, so these are exactly two simple poles per displayed fundamental cell. They also show there is no additional period: a period must permute these two pole classes; exchanging them would shift by $2K$ modulo the displayed lattice, which reverses the function's sign and residues rather than preserving it.

For any finite $w$, integrate $s^{\prime }(z)/(s(z)-w)$ around a fundamental parallelogram avoiding its zeros and poles. The opposite edges cancel by periodicity, so the number of zeros of $s-w$ equals the number of poles, namely two. Therefore every value has two inverse images modulo periods, counted with multiplicity. The value infinity likewise has the two simple poles as preimages. Usually the finite-value preimages are distinct; the involution $z\mapsto 2K-z$ interchanges them. At $w=\pm 1,\pm 1/k$, a critical point supplies one double inverse image instead. For example $s(K)=1$, $s^{\prime }(K)=0$, $s^{\prime \prime }(K)=-(1-k^2)\ne 0$. Thus the printed “exactly two” requires the standard multiplicity convention; it would be false if interpreted as two distinct solutions at these branch values of a holomorphic map.

Now choose $z_0=K$ and put $f(z)=s(z+K)$. Since $s(2K-z)=s(z)$, $f$ is even, elliptic for $\Lambda$, and has degree two. The Weierstrass elliptic function for the same lattice is even and has one double pole per cell, hence also degree two. Its generic fibers are precisely $\{z,-z\}$. Consequently an even $f$ is constant on those fibers and descends to a meromorphic function $R$ of $\wp$. At the four fixed classes of $z\mapsto -z$, the even local Laurent expansion makes this descent meromorphic in the squared local coordinate. A meromorphic function on the Riemann sphere is rational, so $f=R(\wp )$. Generic map degrees multiply:Thus $R$ has degree one and the degree-two even elliptic function conclusion isThe determinant condition follows because $f$ is nonconstant. This proves the existence requested, with the explicit allowable shift $z_0=K$.

True. Label the pentagon polygon vertices clockwise from the upper left asUse $a$ for the upward single arrows and $b$ for the rightward double arrows in the original diagram. Their endpoint identifications areAll five polygon vertices are therefore identified. The remaining diagonal polygon edge, from $v_3$ to $v_4$, has its endpoints identified and is the only boundary component of a surface.

The paired polygon edges join half-disc neighbourhoods into disc neighbourhoods. At the common vertex, the five corner sectors join in one chain whose two free ends lie on the unpaired polygon edge: its vertex link in a polygonal schema is a closed interval. Hence that point has a half-disc neighbourhood. This verifies that the quotient space is a surface with boundary, rather than merely counting its CW cells. The paired polygon edges are oppositely traversed around the polygon, so this surface is orientable.

Its CW cells comprise one vertex, three edges (the $a$ pair, the $b$ pair, and the free polygon edge), and one face. ThereforeThe classification of compact surfaces identifies it as a torus with an open disc removed.

Every complete polygon edge pairing gives a compact connected surface with empty manifold boundary. Here is the local reason at the vertices. Replace each polygon corner by a small link interval. The polygon edge pairings identify all endpoints of these link intervals in pairs. In a given vertex equivalence class, the resulting graph is connected: the endpoint identifications generating that vertex equivalence class also connect its corner intervals. Every link graph vertex has degree of a vertex two, so the graph is a circle. Its topological cone is a disc, giving the required local surface model. Edge and face CW cell interiors have the usual disc neighbourhoods as well. A finite triangulation compatible with the polygon edge pairings realizes the result as a finite simplicial complex, giving a Hausdorff, second-countable space.

For every such surface, $\chi =V-2\ge -1$. If it is orientable, the classification of compact surfaces gives $\chi =2-2g$, so $g=0$ or $1$. If it is a nonorientable surface, write $k\ge 1$ for its nonorientable genus; then $\chi =2-k\ge -1$ gives $k\le 3$. The exhaustive list is therefore

To prove that every listed possibility actually occurs, the following six-letter boundary words specify complete pairings of a hexagon:The vertex counts follow by matching starts and ends of equally directed arrows, just as in part (b). The first and third words pair every letter with its inverse, giving orientable surfaces. Each other word contains a letter twice with the same sign, giving a nonorientable surface: an orientation on the face would induce equal, rather than opposite, boundary directions along that glued pair. With the displayed Euler characteristics, the classification of compact surfaces identifies the examples exactly as listed. All five types occur, and no others occur.

We construct both the required triangulation and its polygonal schema. Subdivide the square $Q=[0,1{]}^2$ into a $4\times 4$ grid. Insert the centre of each small square and join it by line segments to its four corners, making four triangles per small square. Identify opposite sides of $Q$ by translations: $(0,t)\sim (1,t)$ and $(t,0)\sim (t,1)$. This is the standard quotient space presentation of the torus.

The grid and centre vertices, with these identifications, define an abstract simplicial complex $K_1$. Every triangle contains its own small-square centre and two consecutive grid vertices, so no triangle has identified vertices and no distinct triangles acquire the same vertex set. Only intended common sides or vertices are shared. Thus this is a genuine finite triangulation, whose geometric realization is homeomorphic to $T^2$. There are $32$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic zero.

Choose the simplex $\tau$ represented in the interior of $Q$ by the triangle with verticesRemove its relative interior. Before identifying the square's sides, the remainder is a closed disc with a triangular hole, hence an annulus. Cut this annulus along the line segment from $(0,0)$ to $(1/4,1/4)$, treating the two banks as distinct boundary arcs. The cut space is a closed disc; traversing its boundary follows the four outer edges, one bank, the three inner edges, and the other bank. It therefore has nine marked boundary sides.

A homeomorphism from this cut disc to a convex nine-sided polygon can be chosen to carry its nine marked sides to the nine polygon sides. One justification is to choose a boundary homeomorphism with the prescribed marked vertices and extend it over the disc by radial extension after disc parametrizations. Transport the identifications to this convex polygon. With labels recording traversal direction, its boundary word isIdentify equally labelled sides in their arrow directions: the two $a$ sides and two $b$ sides restore the torus pairings, while the two $c$ sides undo the cut. Leave $d,e,f$ unpaired. Their union is the triangular boundary of the removed simplex. The resulting quotient space is exactlyThis is the polygonal presentation of a punctured surface. The word supplies all nine sides and every gluing; inverse exponents reverse the boundary traversal direction.

UseA valid polygon is a finite list of at least three vertices in boundary order, with the final vertex connected back to the first. The first vertex is not repeated at the end. A simple polygon also has no intersecting nonadjacent edges. Numerical routines below use finite coordinates; NaN and infinite coordinates are outside this representation's intended domain.

First include the closing edge exactly once:The recursive version and right fold both sum the Euclidean distances between consecutive vertices, including the last-to-first pair. They take linear time. The empty-list convention gives zero, although an empty list is not a valid polygon.

For five points, partition the unit square into four squares of side $1/2$. Assign boundary points to just one neighbouring piece. By the pigeonhole principle, two of the points lie in one piece, whose diameter is its diagonal, $\sqrt{(1/2{)}^2+(1/2{)}^2}=\sqrt{2}/2$.

For eight points, use seven smaller-diameter pieces rather than eight rectangles. Start with the convex pentagonTake its four reflected copies at the corners of the square. The remaining pieces are the left triangle with vertices $(0,1/4),(1/4,1/2),(0,3/4)$, its right-hand reflection, and the central diamond with vertices $(1/4,1/2),(1/2,1/4),(3/4,1/2),(1/2,3/4)$. These seven polygons cover the square with disjoint interiors. To verify coverage directly, for $0\le y\le 1/4$ the two bottom pentagons split the horizontal section at $x=1/2$. For $1/4\le y\le 1/2$, the five successive sections have endpoints $0,y-1/4,3/4-y,y+1/4,5/4-y,1$ and belong respectively to the left triangle, bottom-left pentagon, diamond, bottom-right pentagon and right triangle. Reflection gives the top half.

The diameter of a finite convex hull is the greatest distance between its vertices: if two points are convex combinations of vertices, their difference is a convex combination of vertex differences, and the triangle inequality bounds its norm by the largest such distance. Checking the five vertices of $P$ gives diameter $\sqrt{(1/2{)}^2+(1/4{)}^2}=\sqrt{5}/4$; reflections preserve that value. Each remaining triangle and the diamond has diameter $1/2$, which is smaller. After assigning shared boundaries to one piece, eight points in these seven pieces force a pair in a common piece. Hence some pair is at distance at most $\sqrt{5}/4$.

Both arguments are instances of a pigeonhole distance bound from a partition.

A polygon with five sides and five vertices. A regular pentagon has equal sides and angles; its symmetry group is the order-ten dihedral group, consisting of five rotations and five reflections. Each reflection axis passes through a vertex and the midpoint of the opposite edge.

For a simple polygon whose vertices are lattice points in ${\mathbb{Z}}^2$, its area is $A=I+B/2-1$, where $I$ counts interior lattice points and $B$ counts boundary lattice points.

Positive real number lengths $a_1,\dots ,a_k$, with $k\ge 3$, are the sides of a nondegenerate polygon exactly when $2{\max }_i{a}_i<{\sum }_i{a}_i$. Necessity follows from the triangle inequality: the longest side must be shorter than the path formed by the remaining sides.

For $n\ge 3$ independent lengths with the uniform distribution on $[0,1]$, the probability of forming a nondegenerate polygon is $1-1/(n-1)!$. The necessary and sufficient condition is that every length is less than the sum of the others. The failure events are disjoint except on sets of probability zero. The sum of $n-1$ lengths has probability density function $s^{n-2}/(n-2)!$ on $0<s<1$, by repeated convolution of probability densities. A specified rod is too long with probability ${\int }_0^1(1-s)s^{n-2}/(n-2)!ds=1/n!$; summing over the $n$ rods proves the formula. For three and four rods the probabilities are $1/2$ and $5/6$.

A boundary point of the source half-plane that maps to a polygon vertex. If that corner has angle $\pi \alpha$, the Schwarz-Christoffel mapping has local behavior $F(z)-F(a)\sim C(z-a{)}^{\alpha }$ at its prevertex, so the derivative has exponent $\alpha -1$. Infinity may also be a prevertex, expressed in the local coordinate $1/z$.