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If a domain of finite hyperbolic area $A$ contains a hyperbolic disc of radius $r$ in curvature $-1$, then $2\pi (\cosh r-1)\le A$, hence $r\le \mathrm{arcosh}(1+A/(2\pi ))$. For an interior point of a hyperbolic triangle, the open disc with radius equal to its distance from the boundary lies inside the triangle. Its area is at most $\pi$ by Gauss-Bonnet theorem, giving the elementary bound $r\le \mathrm{arcosh}(3/2)$; this need not be the sharp triangle inradius bound.

Two corresponding side lengths and an equal nonincluded angle need not determine a hyperbolic triangle. Set $A=\pi /6$, $\cosh a=3/2$, $\cosh b=2$. The hyperbolic law of cosines has two distinct positive solutions $c=\ln (3\pm \sqrt{2})$. The two triangles share $a,b,A$ but have different third sides. An included angle instead determines the third side by the cosine rule.

Three corresponding side lengths determine a hyperbolic triangle up to an isometry of the hyperbolic plane. First match an ordered pair of vertices by an isometry. The third vertex lies at the intersection of two prescribed hyperbolic distance circles. There are at most two choices, interchanged by reflection in the geodesic through the matched pair; either choice can therefore be matched.

For curvature minus one, no hyperbolic triangle contains an open hyperbolic disc of radius greater than $\frac{1}{2}\log 3$. Move the disc centre to zero in the Poincare disc model and enclose the triangle in an ideal triangle with the same radial vertex directions. One gap between ideal directions is at least $2\pi /3$. Its opposite geodesic is at distance at most $2\mathrm{artanh}(2-\sqrt{3})=\frac{1}{2}\log 3$ from zero. The symmetric ideal hyperbolic triangle attains the bound as its inradius.

For curvature $-1$, the Poincare disc model has metric and area elementThe given minimizing-diameter property makes the distance from zero to Euclidean radius $R$ equal toRadial symmetry therefore identifies the whole hyperbolic circle in the Poincare disc. Integrating its enclosed hyperbolic area gives

For a geodesic hyperbolic triangle with angles $\alpha ,\beta ,\gamma$, the Gauss-Bonnet theorem states $A=\pi -(\alpha +\beta +\gamma )$, in particular $A\le \pi$, with strict inequality for ordinary finite vertices. Let $d$ be the distance of the interior point $P$ to the nearest side. The open disc of radius $d$ about $P$ is contained in the triangle: a path from $P$ to an exterior point must first cross the boundary at distance at least $d$. A hyperbolic isometry carries its center to zero, so its area is the area already calculated. ConsequentlyThis uses hyperbolic area bounds an inscribed disc and establishes the requested bound without assuming that $P$ is an incenter. A sharper universal triangle bound is possible, but is not needed here.

Use an oriented geodesic hyperbolic $2s$-gon $D$ with paired edges of equal length. Identify each pair by a length-preserving map reversing the boundary directions, to make an oriented base surface $M_0$ with possibly singular vertices. Assign to the $s$ edge pairs generators $A_1,\dots ,A_s$ of $T$, with inverse labels for reversed crossings. For each $g\in T$, take a copy $D_g$ and glue its positively labelled edge of type $k$ to the corresponding mate in $D_{g{A}_k}$. The generators make the glued surface $M$ connected. Left multiplication $D_g\mapsto D_{hg}$ commutes with these right-labelled gluings and gives the $T$ action with quotient $M_0$.

Existence for an arbitrary finite group can even be arranged with no cone singularities. If $T$ has $d$ generators, take $h\ge \max \{2,d\}$ and a regular hyperbolic polygon with $4h$ edges, all angles $\pi /(2h)$, paired with side word ${\prod }_{i=1}^h{a}_i{b}_i{a}_i^{-1}{b}_i^{-1}$. Its vertices form one cycle of angle $2\pi$, giving a smooth base of genus $h$. Label the $a_i$ by the generators of $T$, padded by identities, and every $b_i$ by the identity. The single vertex word is ${\prod }_i[A_i,1]=1$, so every lifted vertex also has angle $2\pi$. The resulting connected ordinary normal covering map has degree $|T|$ and deck group $T$.

This is the paired-polygon construction of finite surface covers. Interior and open-edge points are regular hyperbolic points; possible singularities occur only at vertices. When the base has cone points, the map is an orbifold covering and may be branched as a map between the underlying topological surfaces. It should not be called an unbranched manifold covering at those branch points.

A vertex cycle of a paired polygon is the collection of vertices identified to one point on $M_0$. Let $r$ be the number of cycles, ${\theta }_j$ the sum of the polygon angles in cycle $j$, and $C_j\in T$ the gluing word recorded on going around it. If $q_j=|C_j|$ is the order of that element, a lift closes after $q_j$ turns. It has angle $q_j{\theta }_j$, and there are $|T|/q_j$ such lifted vertices. The base and lifted cone orders, when they are integer orbifold orders, areAn orbifold homomorphism requires $q_j\mid m_j$. The lifted point is smooth exactly when $q_j{\theta }_j=2\pi$, equivalently $q_j=m_j$ in that case. For arbitrary cone metrics the angle formulas remain valid even when an angle does not correspond to an integer orbifold order.

Counting faces, edges and vertices gives the Euler characteristic of a paired-polygon surface cover:These are ordinary topological Euler characteristics, rather than orbifold Euler characteristics. If the lifted surface is smooth, the equivalent orbifold formula isFor a closed oriented smooth surface, $\chi =2-2g$ defines its genus. These formulas also follow from the Riemann-Hurwitz formula by subtracting the ramification deficits $|T|-|T|/q_j$ over each base vertex.

For $U\le T$, the quotient map $M\to M_1=U\backslash M$ is an ordinary normal covering if and only if $U$ acts freely. All possible point stabilizers in this construction are conjugates of $⟨C_j⟩$, so the concrete necessary and sufficient condition isThis is the freeness criterion for a subgroup of a polygon-cover deck group. Under it the deck group is $U$, the degree is $|U|$, andThere is no requirement $U◃T$ for this map. That different condition characterizes normality of the intermediate cover $M_1\to M_0$. If fixed points are retained and one instead speaks of an orbifold covering, the quotient is regular as an orbifold cover, but ordinary Euler characteristic need not divide by $|U|$; orbifold Euler characteristic does.

For the low-genus examples takeLet $U_1$ be a nonzero-vector stabilizer and $U_2$ a plane stabilizer. Their actions have seven points, so $|U_1|=|U_2|=24$ and $[T:U_i]=7$. They are almost conjugate subgroups: the numbers of fixed nonzero vectors for a matrix $A$ and its dual action $A^{-T}$ are both $2^{\dim \ker (A-I)}-1$. Thus the two coset permutation characters agree, which is equivalent to Gassmann equivalence. They are not conjugate: a point stabilizer has a common fixed nonzero vector, whereas a plane stabilizer fixes no nonzero vector globally. In particular, since $7\nmid 24$, neither subgroup contains any nonidentity element of a cyclic subgroup of order seven. All cone monodromies of exact order seven therefore act freely on the intermediate covers.

For genus three, start with a hyperbolic triangle whose three angles are $\pi /7$ and double it across one edge. The resulting quadrilateral $D$ has anglesThe side pairings produce three base vertex cycles, each of total angle $2\pi /7$, and an underlying sphere. Choose generators $A,B$ with $|A|=|B|=|AB|=7$; the three vertex words are $A,B,(AB{)}^{-1}$. Hence $s=2$, $r=3$ and all $q_j=7$. The full cover is smooth, andThe hyperbolic polygon area of $D$ is $8\pi /7$, so the seven-sheeted surfaces have area $8\pi$, agreeing with Gauss-Bonnet. Their equal spectra follow from the Sunada theorem. For complete numerical generator data, one possible pair isThese generate all of $T$ and have the required orders. This construction establishes the requested isospectrality; it does not establish nonisometry. In the constant-curvature triangle example a lifted reflection can identify the point and plane covers, as in reflection intertwining of triangle-cover coset actions. Breaking that symmetry with a generic metric gives nonisometric genus-three surfaces, but then the metric need not have constant curvature. Those are distinct assertions.

For genus four, use the cone-torus construction of genus-four Sunada surfaces. Take a regular hyperbolic quadrilateral with four anglesand pair opposite sides in the usual torus pattern $a,b,a^{-1},b^{-1}$. All four vertices belong to one cycle, whose total angle is $2\pi /7$; the underlying base is a torus with one cone point. Choose $A,B$ of order four generating $T$, with commutator $[A,B]$ of order seven. For example,Only the commutator is a cone monodromy, so the order-four side labels do not create cone points of order four. Here $s=2$, $r=1$, $q_1=7$, givingThe quadrilateral has area $12\pi /7$, and each seven-sheeted smooth quotient has area $12\pi$. Again the order-seven cone stabilizer meets neither $U_i$ nontrivially, so the metrics are smooth and the Sunada theorem gives identical spectra. The required group and subgroup orders are therefore 168 and 24 in both constructions; the genus-three side-generator orders are 7 and 7, and the displayed genus-four side-generator orders are 4 and 4, with cone-word order 7 in both cases.

In the upper half-plane model $H=\{(x,y):y>0\}$ the hyperbolic metric isThe length of a differentiable curve is $\int \sqrt{{\dot{x}}^2+{\dot{y}}^2}dt/y$. The area element is the square root of the metric determinant times $dxdy$, so the hyperbolic area of a measurable region $S$ is ${\int }_S{y}^{-2}dxdy$.

The Gauss-Bonnet theorem for a geodesic hyperbolic triangle of angles $\alpha ,\beta ,\gamma$ in curvature $-1$ gives area $\pi -\alpha -\beta -\gamma$; an ideal vertex has angle zero. The horizontal edge in the present region is not a geodesic, so it is simpler to use the area element directly:The open boundary does not affect the integral because its area is zero.

Suppose an open hyperbolic disc centred at $X$ is contained in a hyperbolic triangle. Move $X$ to zero by an isometry. Extend the radial geodesics from zero through the three vertices to ideal points $P_1,P_2,P_3$. Because zero lies inside the original triangle, the three vertex directions are not contained in a semicircle. The ideal hyperbolic triangle with those endpoints therefore contains zero: otherwise a separating geodesic through zero would put all three endpoints in one semicircle. It also contains each finite vertex, which lies on the segment from zero to the corresponding ideal endpoint. Geodesic convexity then makes it contain the original triangle.

The three angular gaps between successive ideal endpoints sum to $2\pi$, and each is at most $\pi$. One has size at least $2\pi /3$, so its half-angle $\alpha$ is at least $\pi /3$. The distance in part (iii) decreases as $\alpha$ increases. Thus the corresponding side of the ideal triangle is at distance at mostfrom zero. If the disc radius exceeds this distance, it contains a point on that side and points beyond it, and cannot be contained in the ideal triangle, hence cannot be contained in the original one. This proves the hyperbolic triangle inradius bound:The symmetric ideal triangle has equality, explaining the constant. Finite triangles have strictly smaller inradius.

Subgroups $U,V$ of a finite group $T$ have Gassmann equivalence whenSumming over the conjugacy classes gives $|U|=|V|$. Equivalently, their coset permutation representations have the same character of a representation. In particular, almost conjugate subgroups need not be conjugate.

Put $p=\mathrm{ord}A$, $q=\mathrm{ord}B$, and $r=\mathrm{ord}(AB)$. Begin with the sphere punctured at three points. Its fundamental group is $⟨a,b,c\mid abc=1⟩$, and the epimorphism sending $a,b,c$ to $A,B,(AB{)}^{-1}$ defines a connected regular $|T|$-sheeted covering space. Around the punctures the monodromies have cycles of lengths $p,q,r$. Fill each lifted puncture by a disk, using the local branched covering models $z\mapsto z^p$, $z\mapsto z^q$ and $z\mapsto z^r$. This produces a closed oriented surface $M$, with $T$ acting by homeomorphisms that are smooth in these local models. This construction works even if one of $p,q,r$ is one.

Triangulate the base sphere with the three marked points as vertices and two faces. Lifting this triangulation givesso the Euler characteristic isIn the hyperbolic case $1/p+1/q+1/r<1$, this is equivalently the triangle cover construction for Sunada surfaces. Take the hyperbolic triangle groupand its epimorphism $\theta :\Delta \to T$ with the same images. All finite-order elements of $\Delta$ are conjugate to powers of its three elliptic generators. Because their images have the same exact orders, $K=\ker \theta$ is torsion-free and $M=K\backslash {\mathbb{H}}^2$. It can be assembled from $2|T|$ hyperbolic triangles of angles $\pi /p,\pi /q,\pi /r$. Both examples below belong to this hyperbolic case.

For $M_1=U\backslash M$ and $M_2=V\backslash M$ to be smooth surfaces normally covered by $M$, the subgroup actions must be free. The precise condition is that neither subgroup meet any conjugate in $T$ of a nonidentity power of $A$, $B$ or $AB$. These are exactly the possible point stabilizers. Then $M\to M_1$ and $M\to M_2$ are regular coverings, with deck transformation groups $U,V$ respectively. No normality of $U,V$ in $T$ is needed: normality would concern the further covering of the base orbifold, not these coverings from $M$. For a free subgroup $W$,

Choose any $T$-invariant smooth Riemannian metric on $M$, for example by averaging an arbitrary metric. It descends to both quotients. On functions let $R_t$ be the unitary action of $t\in T$, and let $E=e^{-s{\Delta }_M}$ for $s>0$. The averaging operator $P_W=|W{|}^{-1}{\sum }_{w\in W}{R}_w$ projects orthogonally onto the $W$-invariant functions, which identify with functions on $W\backslash M$. The trace formula underlying the Sunada theorem isThe expression $\mathrm{Tr}(R_w{e}^{-s{\Delta }_M})$ is a class function, because ${\Delta }_M$ commutes with $T$. Thus Gassmann equivalence makes the two sums identical. Alternatively, apply the same averaging formula to each finite-dimensional Laplacian eigenspace: its invariant dimension is $|W{|}^{-1}{\sum }_{w\in W}{\chi }_{\lambda }(w)$. This proves equality of every eigenvalue multiplicity and hence isospectrality of the two quotient metrics.

If $V=tU{t}^{-1}$ for some $t\in T$, the action of $t$ gives an isometry between the quotients for every such metric. Nonconjugacy is therefore necessary. To make it sufficient, choose the quotient orbifold metric generically so that its regular locus has no nonidentity local isometries; this is the equivariant, or orbifold, version of the Sunada local isometry lemma. Smooth perturbations on regular coordinate balls provide this genericity while their lifts remain smooth on $M$. Any isometry between $M_1$ and $M_2$, when composed locally with their projections, would then identify the same points of the base regular locus. By continuation it is an isomorphism of their coverings of that locus, and the classification of connected covering spaces forces $U,V$ to be conjugate in $T$. Thus nonconjugate subgroups give non-isometric isospectral quotients for a suitable generic invariant metric. The genericity belongs to the base regular locus; the covering metric on $M$ itself necessarily has the prescribed deck symmetries.

For genus two, an explicit choice is the following permutation group on twelve labels, with permutation products composed from right to left:These parameters are those in Kang's genus-two construction. They admit a direct finite verification. Starting with $T_0=\{1\}$ and iterating $T_{j+1}=T_j\cup T_{j}A\cup T_{j}B$, the successive sizes are $1,3,7,13,23,39,61,81,93,96$, after which the set is unchanged. Thus $|T|=96$, and direct multiplication gives $(p,q,r)=(3,3,6)$ and $|U|=|V|=8$. The conjugacy-class triples $(|C|,|C\cap U|,|C\cap V|)$, obtained as $C=\{tx{t}^{-1}:t\in T\}$, areThis proves Gassmann equivalence. The subgroup $V$ fixes the labels $5,11$, whereas $U$ has no common fixed label, so they cannot be conjugate even in the full symmetric group. Both are groups of order eight, so they meet no nonidentity order-three element. The only order-two power of $AB$ iswhich is central in $T$ and lies in neither subgroup. This also excludes its conjugates and proves freeness. The quotient Euler characteristics and genera are thereforeThe covering surface $M$ has genus nine.

For genus three, use $T={\mathrm{GL}}_3({\mathbb{F}}_2)$, of order $(8-1)(8-2)(8-4)=168$. Let $U$ stabilize a nonzero vector and let $V$ stabilize a two-dimensional subspace. Both have index seven and order twenty-four. An element $t$ fixes $2^{\dim \ker (t-I)}-1$ nonzero vectors. Hyperplanes correspond to nonzero dual vectors over ${\mathbb{F}}_2$, so the number fixed in the dual action is $2^{\dim \ker (t^T-I)}-1$, the same number. Equality of these permutation characters proves that the point and hyperplane stabilizers are almost conjugate.

They are not conjugate. In coordinates where $U$ fixes $e_1$, its elements include all matrices $\left(\begin{array}{cc}1& \alpha \\ 0& C\end{array}\right)$ with arbitrary row $\alpha$ and $C\in {\mathrm{GL}}_2({\mathbb{F}}_2)$. An invariant plane containing $e_1$ would project to a line invariant under every $C$, which is impossible. A plane not containing $e_1$ would be the graph of a linear functional on the quotient, and arbitrary $\alpha$ would move that graph. Thus $U$ stabilizes no plane, whereas every conjugate of $V$ does.

Choose the concrete generatorsOver ${\mathbb{F}}_2$, each of $A,B,AB$ has order seven. The same closure iteration has sizes $1,3,7,15,31,59,103,144,168$, verifying that $A,B$ generate $T$. Since seven does not divide $24$, neither stabilizer contains a nonidentity power of these generators, or any conjugate of such a power. We may therefore use $(p,q,r)=(7,7,7)$ in the triangle cover construction for Sunada surfaces. The quotient calculation isHere the covering surface $M$ has genus forty-nine. Choosing the generic invariant metrics described above completes both requested pairs.

An ideal hyperbolic triangle has its three vertices at infinity, with sides given by geodesics. In curvature $-1$ its angles are zero and its area is $\pi$, using the hyperbolic triangle area formula $\pi -(\alpha +\beta +\gamma )$.

To calculate the area of a hyperbolic disc, use the Poincare disc model centered at the origin. Hyperbolic radial distance is $\rho ={\int }_0^{R}2dr/(1-r^2)=2\mathrm{artanh}R$, so its Euclidean radius is $R=\tanh (\rho /2)$. The metric area density is $4rdrd\theta /(1-r^2{)}^2$, givingEvery hyperbolic triangle, including ones with ideal vertices, has area at most $\pi$. But $A(2)=2\pi (\cosh 2-1)>\pi$. A hyperbolic triangle is geodesically convex, so if it contains a complete hyperbolic circle, it also contains the enclosed hyperbolic disc, its geodesic convex hull. This would violate the area comparison. Therefore no hyperbolic triangle contains a complete circle of radius two.

In the hyperbolic law of cosines, the angle opposite $a$ being at least $\pi /2$ makes its cosine nonpositive. ThusSince $\cosh$ is increasing on $[0,\infty )$, the opposite side is longest: $\boxed{a\ge b,a\ge c}$. For a nondegenerate hyperbolic triangle both comparisons are strict.

At an interior point $w$ of the geodesic segment from $z_1$ to $z_2$, the two angles made with the segment toward $w^{\prime }$ add to $\pi$. At least one is at least $\pi /2$. In the corresponding hyperbolic triangle, part (a) says that its opposite side $w^{\prime }{z}_i$ is at least the adjacent side $w^{\prime }w$. ThereforeIf $w$ is an endpoint the assertion is immediate. The same inequality also holds when $w^{\prime }$ lies on the complete geodesic through $z_1,z_2$: parameterize that geodesic by arclength and use the elementary inequality $|t-s|\le \max (|t-a|,|t-b|)$ for $a\le s\le b$. This extension will allow repeated use of the result in part (c).

Consider the filled hyperbolic triangle with vertices $A,B,C$. Every point $P$ is on some geodesic segment $Aw$ with $w\in BC$: for instance, in the Beltrami-Klein model these are the ordinary straight segments filling a convex triangle. For any other point $Q$ of the triangle, apply part (b) first on $Aw$ and then on $BC$:Applying the same argument to $Q$, with each vertex held fixed, bounds every term on the right by the maximum of the three side lengths. The endpoints of a longest side attain that bound. Thus the diameter is exactly the longest side length:

In the curvature $-1$ normalization, the hyperbolic triangle area is its angle defect:Join one vertex of a convex geodesic polygon to its nonadjacent vertices. This gives $n-2$ hyperbolic triangles with disjoint interiors. Their angles at every polygon vertex sum to that vertex's interior angle. Adding the hyperbolic triangle areas therefore gives the hyperbolic polygon areaFor a regular hyperbolic polygon with prescribed area, place $n$ equally spaced vertices on a hyperbolic circle of radius $R>0$ and join consecutive vertices by geodesics. Rotations through $2\pi /n$ and reflections in radial lines show that this is a convex regular polygon. The triangle from its centre to a vertex and the midpoint of an adjacent side is right angled, with angles $\pi /n$, $\alpha (R)/2$, and $\pi /2$, and hypotenuse $R$. The angle form of the hyperbolic law of cosines givesConsequently $A(R)=(n-2)\pi -n\alpha (R)$ is continuous and strictly increasing. As $R\downarrow 0$, $\alpha (R)\to \pi -2\pi /n$, so $A(R)\to 0$; as $R\to \infty$, $\alpha (R)\to 0$, so $A(R)\to (n-2)\pi$. The intermediate value theorem proves existence for every permitted $A$. Indeed the required radius isThe argument also proves uniqueness of the radius in this construction; neither endpoint area is attained by a nondegenerate finite-radius polygon.

Because the half-plane metric is a conformal rescaling of the Euclidean metric, its angles are Euclidean angles. For each $y>0$, the hyperbolic lines meeting the imaginary axis at a fixed angle $\alpha$ at $iy$ are the Euclidean circleswhere $Y$ denotes the vertical coordinate. These circles have centres on the real axis and are therefore hyperbolic lines; varying $y$ gives the required collection.

To construct a triangle, put one vertex at $i$ and a second at $it$ on ${\ell }_{+}$, with $t>1$. Draw from these vertices the lines making interior angles $\alpha$ and $\beta$ with ${\ell }_{+}$. Their Euclidean centres and radii areIf their other intersection has angle $\gamma$, the law of cosines in the Euclidean triangle formed by the two centres and that intersection givesEquivalently,The right side is greater than one exactly when $\alpha +\beta +\gamma <\pi$. Since $(t+t^{-1})/2$ ranges continuously and strictly increasingly from $1$ to infinity for $t>1$, such a $t$ exists. The three lines therefore bound a hyperbolic triangle with the prescribed angles.

A hyperbolic triangle is bounded by three hyperbolic geodesic segments or rays. Its vertices may lie in the hyperbolic plane; an ideal vertex is their endpoint on the boundary at infinity. Every ideal angle is zero. Gauss-Bonnet theorem with curvature $-1$ givesso an all-ideal triangle has area $\pi$.

For fixed admissible angles, hyperbolic trigonometry determines all three side lengths from the angles, for exampleThus two such triangles are congruent. An orientation-preserving isometry can send one chosen vertex and oriented tangent to the corresponding data of the other, and the determined side lengths and angles then send the entire triangle to it. Since the orientation-preserving isometry group is $PS{L}_2(\mathbb{R})$, represented by $S{L}_2(\mathbb{R})$, the stated action is transitive.