Divide the two identities and apply the chain rule. The transformed length integrand is exactly the original. A unit rotation adds no change of modulus.

As \(r\to1^-\), the distance tends to infinity. For \(r=1/2\), the distance is \(\log3\).

Let \(\rho(t)=|z(t)|\). Almost everywhere \(|\rho\prime(t)|\le|z\prime(t)|\). Therefore

The radial segment attains equality. Moving endpoints by an isometry proves the corresponding geodesic is a shortest path.

It is twice the distance from zero to one-half, consistent with additivity along the diameter.

The half-plane metric is \(ds=|dw|/\operatorname{Im}w\). Integrating along the vertical geodesic gives:

Let r be the height ratio between successive markers. Equal logarithmic distances mean the same ratio, so \(r^3=9\). The heights are \(1,9^{1/3},9^{2/3},9\). Each segment has length \(\log9/3\). Equally spaced Euclidean heights would have unequal hyperbolic lengths.

Move its center to zero. Its Euclidean radius is \(r=\tanh(R/2)\). Parametrize the circle by \(z=re^{it}\) and integrate:

Draw diagonals from one vertex, giving n minus two nonoverlapping triangles. Their angle sums combine into the polygon angle sum. Add their defects:

In particular an ideal quadrilateral has area two pi. A nondegenerate four-right-angle rectangle would have zero area, which is impossible.

Hence \(\theta<\pi/2\). Equal perpendicular legs force equal summit angles by reflection in the perpendicular bisector of the base.

Move the center to zero and integrate the area element out to Euclidean radius r.

Use the derivative of sinh at zero, or its power series. The limit explains why very small regions look Euclidean even though the global geometry is different.

Approach. Normalize the figure using an isometry, then apply the definition.

Since \(|z(t)-a|=r\) and the argument traverses a full turn, the center is \(a\) and the Euclidean radius is \(r\) (assuming \(r>0\)). This is not, in general, its hyperbolic center or radius.

Approach. Normalize the figure using an isometry, then apply the definition.

Writing \(a=u+iv\) gives \(x(t)=u+r\cos t\) and \(y(t)=v+r\sin t\).

Approach. Normalize the figure using an isometry, then apply the definition.

Both coordinates are infinitely differentiable and \(z'(t)=ire^{it}\) has constant modulus \(r>0\), so it is a regular smooth parametrization. For \(r=0\) it is constant and smooth but not regular. To regard it as a curve inside the hyperbolic disk, require \(|a|+r<1\).

Approach. Normalize the figure using an isometry, then apply the definition.

For \(0\le r<1\), \(d(0,r)=\log((1+r)/(1-r))\). The numerator tends to \(2\) and the positive denominator tends to zero, so the distance tends to \(+\infty\). The boundary is not part of the hyperbolic plane.

Approach. Normalize the figure using an isometry, then apply the definition.

Concatenate the minimizing segments from \(z_1\) to \(z_2\) and from \(z_2\) to \(z_3\). Its length is \(d(z_1,z_2)+d(z_2,z_3)\). The minimizing segment from \(z_1\) to \(z_3\) has no greater length, proving the inequality. Piecewise smooth curves are allowed, or the corner can be approximated by smooth curves.

Approach. Normalize the figure using an isometry, then apply the definition.

The geodesics are vertical rays and upper semicircles centered on the real axis. These are exactly the clines orthogonal to the boundary. A conformal disk-to-half-plane Möbius map carries disk geodesics to these curves.

Approach. Normalize the figure using an isometry, then apply the definition.

It is the angle of parallelism. If the perpendicular distance is \(d\), then \(\theta=2\arctan(e^{-d})\). The two limiting parallel rays make equal angles with the perpendicular.

Approach. Normalize the figure using an isometry, then apply the definition.

Apply a disk isometry taking \(z_1\) to \(0\) and \(z_2\) to \(r>0\). It takes the ordered ideal endpoints to \(-1,1\). With the book's convention, \[[0,r,1,-1]=\frac{0-1}{0+1}\frac{r+1}{r-1}=\frac{1+r}{1-r}.\] Both distance and cross ratio are invariant. Therefore \(d(z_1,z_2)=\log[z_1,z_2,q_2,q_1]\). Reversing the endpoint order inverts the ratio; its logarithm changes sign, so the ordering matters.

Approach. Normalize the figure using an isometry, then apply the definition.

Set \(\rho=|(z_2-z_1)/(1-\bar z_1z_2)|\). The identity \[1-\rho^2=\frac{(1-|z_1|^2)(1-|z_2|^2)}{|1-\bar z_1z_2|^2}\] shows \(\rho\to1\), since the denominator is bounded below by \((1-|z_1|)^2>0\). Hence \(\log((1+\rho)/(1-\rho))\to\infty\). Fixing \(z_1\) is essential; two points approaching the same boundary point together may stay a bounded distance apart.

Approach. Normalize the figure using an isometry, then apply the definition.

Normalize their geodesic to the real diameter and let \(-1<x_1<x_2<x_3<1\). Define \(F(x)=\log((1+x)/(1-x))\), whose derivative is \(2/(1-x^2)>0\). Integrating the line element gives \(d(x_i,x_j)=F(x_j)-F(x_i)\) for \(i<j\). The differences telescope, so \(d(z_1,z_3)=d(z_1,z_2)+d(z_2,z_3)\).

Approach. Normalize the figure using an isometry, then apply the definition.

Move the center to \(0\). Its cycle becomes \(|z|=r\), and every point has distance \(R=\log((1+r)/(1-r))\). Conversely this strictly increasing function of \(r\) takes the value \(R\) only at \(r=\tanh(R/2)\). Thus the entire distance locus is that circle. Transfer back by the inverse isometry.

Approach. Normalize the figure using an isometry, then apply the definition.

Move the endpoints to \(0,\infty\) in the half-plane. The geodesic is the imaginary axis and one branch of the hypercycle is a ray \(z=te^{i\phi}\), \(t>0\), with fixed \(0<\phi<\pi\). Dilations \(z\mapsto cz\), \(c>0\), preserve the axis and act transitively on this ray. They preserve perpendiculars and length, so the perpendicular distance is constant.

More explicitly the perpendicular through \(te^{i\phi}\) is the circle \(|z|=t\), and its arc to \(it\) has length \(|\int_\phi^{\pi/2}\csc u\,du|=|\log\tan(\phi/2)|\). The case \(\phi=\pi/2\) is the axis itself.

Approach. Normalize the figure using an isometry, then apply the definition.

On an interval avoiding the pole, the real-variable product and quotient rules apply equally to complex-valued functions, by applying them to real and imaginary parts. Thus \[w' =\frac{az'(cz+d)-cz'(az+b)}{(cz+d)^2}=\frac{ad-bc}{(cz+d)^2}z'=T'(z)z'.\] If a curve passes through infinity, use the local coordinate \(1/z\) there instead of this finite chart.

Approach. Normalize the figure using an isometry, then apply the definition.

Move the vertex to \(0\). The sides become radial rays with arguments \(-\alpha\) and \(\alpha\), where \(0<\alpha<\pi/2\). Reflection in the intervening diameter exchanges the sides and is an isometry, so its interior ray is equidistant from them.

For the converse, use the perpendicular-distance formula to a diameter: for \(z=re^{i\theta}\) it is \(\operatorname{arsinh}(2r|\sin(\theta-\beta)|/(1-r^2))\) for the diameter at angle \(\beta\). Equality for \(\beta=\pm\alpha\) gives \(\sin^2(\theta-\alpha)=\sin^2(\theta+\alpha)\), hence \(\sin2\theta\sin2\alpha=0\). Within \(-\alpha<\theta<\alpha\), only \(\theta=0\) occurs. The relevant feet lie on the side rays. This proves the interior locus and uniqueness.

Approach. Normalize the figure using an isometry, then apply the definition.

Follow the internal bisector from vertex \(A\) to its intersection \(P\) with side \(BC\). Along this segment let \(f(X)=d(X,AB)-d(X,BC)\), using supporting lines. Near \(A\), \(f<0\); at \(P\), \(f>0\). Continuity gives an interior point \(I\) with equal distances to \(AB\) and \(BC\). The bisector property also gives \(d(I,AB)=d(I,AC)\), so \(I\) lies on all three internal bisectors. Two distinct geodesics meet at most once in the disk, giving uniqueness.

Let their common distance be \(r>0\). The closed metric disk centered at \(I\) of radius \(r\) lies in each of the three side half-planes: a point outside one would force a shorter crossing of its boundary. It is therefore contained in the triangle. Its perpendicular contact points lie on the side segments, and the circle is tangent there.

Approach. Normalize the figure using an isometry, then apply the definition.

If the bisectors concur at \(O\), their locus property gives \(d(O,A)=d(O,B)=d(O,C)\); the distance circle through any vertex passes through all three.

For a counterexample in the upper half-plane, take \(A=-1+i\), \(B=i\), \(C=1+i\). The formula \[\cosh d(u+iv,x+i)=1+\frac{(u-x)^2+(v-1)^2}{2v}\] gives \(d(O,A)=d(O,B)\) only when \(u=-1/2\), and \(d(O,B)=d(O,C)\) only when \(u=1/2\). No \(v>0\) satisfies both. These three points form a nondegenerate hyperbolic triangle: a horizontal Euclidean line is a horocycle, not a geodesic.

Approach. Normalize the figure using an isometry, then apply the definition.

Such a map is a sphere homeomorphism, so it maps the boundary \(\mathbb R\cup\{\infty\}\) onto itself. The images of \(0,1,\infty\) are distinct real extended points. There is a real-coefficient fractional linear map with those three images, obtained by the three-point normalization formula. Uniqueness of a Möbius map on three points makes it the given map.

For real coefficients, \[\operatorname{Im}T(z)=\frac{(ad-bc)\operatorname{Im}z}{|cz+d|^2}.\] Since both imaginary parts are positive, \(ad-bc>0\). Conversely this identity and the inverse show every such map is a bijection of the upper half-plane. Matrix multiplication preserves real entries and multiplies positive determinants; the identity and inverse also have positive determinant. Hence these maps form a group.

Approach. Normalize the figure using an isometry, then apply the definition.

A distance circle of radius \(R\) centered hyperbolically at \(i\) is \(x^2+(y-\cosh R)^2=\sinh^2R\). A horocycle is \(y=1\), or a Euclidean circle internally tangent to \(y=0\). A hypercycle is \(x=y\) with \(y>0\), equidistant from the imaginary axis; its other cline endpoint is \(\infty\). More generally a cline meeting the real boundary at two distinct points nonorthogonally gives a hypercycle.

Approach. Normalize the figure using an isometry, then apply the definition.

A horizontal line \(y=c>0\) is a horocycle based at \(\infty\). The part of a nonhorizontal, nonvertical line in the upper half-plane is a hypercycle: it meets the boundary at a finite point and at \(\infty\), but not orthogonally. A vertical line is a geodesic, the zero-distance limiting case.

Approach. Normalize the figure using an isometry, then apply the definition.

Use \(F(z)=i(1+z)/(1-z)\), with inverse \(F^{-1}(w)=(w-i)/(w+i)\). Since \(\operatorname{Im}F(z)=(1-|z|^2)/|1-z|^2\), this is a bijection from the disk to the upper half-plane. Conjugation \(T\mapsto FTF^{-1}\) sends every disk automorphism to a half-plane automorphism, and the reverse conjugation proves surjectivity between the groups. It also preserves length because \[\frac{|F'(z)|}{\operatorname{Im}F(z)}=\frac{2}{1-|z|^2}.\] Thus both the transformation geometry and its metric correspond.

Approach. Normalize the figure using an isometry, then apply the definition.

Center it at zero, with Euclidean radius \(r=\tanh(R/2)\). Parametrize by \(re^{it}\). Then \[C=\int_0^{2\pi}\frac{2r}{1-r^2}\,dt=\frac{4\pi r}{1-r^2}=2\pi\sinh R.\] The last equality follows from \(2\tanh u/(1-\tanh^2u)=\sinh2u\).

Approach. Normalize the figure using an isometry, then apply the definition.

Lobachevsky's formula gives \(e^{-d}=\tan(\pi/8)=\sqrt2-1\). Therefore \[d=-\log(\sqrt2-1)=\log(1+\sqrt2).\]

Approach. Normalize the figure using an isometry, then apply the definition.

Place its right vertex at \(0\) and the other vertices at \(r\) and \(is\), with \(0<r,s<1\). The disk distance formula implies \[\cosh d(z,w)=1+\frac{2|z-w|^2}{(1-|z|^2)(1-|w|^2)}.\] Hence \[\cosh c=1+\frac{2(r^2+s^2)}{(1-r^2)(1-s^2)}=\frac{(1+r^2)(1+s^2)}{(1-r^2)(1-s^2)}=\cosh a\cosh b.\] This is the Pythagorean relation because it determines the opposite side from the two perpendicular legs. At small lengths, expansion gives \(c^2\approx a^2+b^2\).

Approach. Normalize the figure using an isometry, then apply the definition.

Normalize the horocycle to \(y=h>0\) and one endpoint to \(ih\). Its tangent geodesic is the semicircle centered at \(0\) of radius \(h\), with ideal endpoints \(\pm h\). The horocycle diameter at the other endpoint is vertical, so asymptoticity requires that endpoint to be \(h+ih\) or \(-h+ih\). The horocycle arc length is \(\int_0^h dx/h=1\).

Translations and positive dilations take any normalized arc to the segment from \(i\) to \(1+i\), after choosing its endpoint order. All horocycles can first be normalized this way by a Möbius isometry. Thus the unoriented fundamental arcs are congruent.

Approach. Normalize the figure using an isometry, then apply the definition.

Move the ideal point to infinity. The horocycles are \(y=h_1\) and \(y=h_2\) with \(h_2>h_1\), and the diameters are vertical lines \(x=a,b\). Along either vertical line, \(d=d'=\int_{h_1}^{h_2}dy/y=\log(h_2/h_1)\). The arc lengths are \(s_j=(b-a)/h_j\), so \(s_2/s_1=h_1/h_2=e^{-d}\). Reversing the labels reverses the sign in the exponential.

Approach. Normalize the figure using an isometry, then apply the definition.

A straightedge is a rigid geodesic segment; move it so it passes through two specified points and extend its geodesic. A compass keeps one point a fixed hyperbolic distance from its pivot; rotating it traces a hyperbolic circle. In a disk drawing these instruments do not appear Euclidean-straight or Euclidean-rigid near the boundary: the metric, rather than apparent screen length, defines rigidity.

Approach. Normalize the figure using an isometry, then apply the definition.

Fix an ideal direction, choose a reference horocycle, and move a pen while keeping its Busemann height constant. In upper-half-plane coordinates with that direction at infinity, the height is \(\log y\) and the pen traces \(y=h\). Intrinsically the instrument can follow a family of asymptotic geodesics and maintain equal signed distance from the reference horocycle. This is an idealized geometric construction; the ideal point cannot be used as a physical finite compass pivot.

Approach. Normalize the figure using an isometry, then apply the definition.

Let a carriage travel along a geodesic rail. Attach a fixed-length perpendicular arm and put a pen at its end. As the carriage moves, the pen stays at a fixed signed distance from the rail and traces one hypercycle. The arm length zero gives the rail; choosing the other side produces the other equidistant branch.

Approach. Normalize the figure using an isometry, then apply the definition.

The turn is the angle of parallelism, so \(e^{-d}=\tan(\theta/2)\). Thus \[d=\log\cot(\theta/2),\qquad0<\theta\le\pi/2.\] At \(\theta=\pi/2\) the distance is zero; as \(\theta\to0\), the distance grows without bound.

Approach. Normalize the figure using an isometry, then apply the definition.

Choose four distinct cyclically ordered boundary points and join successive points by geodesics. A diagonal splits the quadrilateral into two ideal triangles. Each has area \(\pi\), so the quadrilateral has area \(2\pi\).

Approach. Normalize the figure using an isometry, then apply the definition.

Triangulate a simple geodesic polygon with \(n\) sides into \(n-2\) triangles. Adding their angle defects gives \[A=(n-2)\pi-\sum_{j=1}^{n}\alpha_j.\] For a quadrilateral this is \(2\pi-\sum\alpha_j\). Interior ideal angles are zero; the formula extends by a limit. The polygon must be simple, with its interior and angle conventions specified.

Approach. Normalize the figure using an isometry, then apply the definition.

Four right angles would give \(A=2\pi-4(\pi/2)=0\). A nondegenerate polygon has positive area, a contradiction. Thus no geodesic rectangle exists.

Approach. Normalize the figure using an isometry, then apply the definition.

No. Its enclosed area is \(A(R)=2\pi(\cosh R-1)\), which tends to infinity with \(R\). Every finite circle can be enlarged. A bounded Euclidean disk drawing does not imply a bounded hyperbolic area.

Approach. Normalize the figure using an isometry, then apply the definition.

After aligning a corresponding vertex and its two rays, compare the two radial distances independently: each is less than, equal to, or greater than its counterpart, giving nine combinations. Both equal gives identical triangles. Both strictly smaller or both strictly larger gives strict containment, contradicting equal angle-defect areas. Opposite strict inequalities gives the crossing configuration and a triangle with angle sum at least \(\pi\), impossible. If exactly one comparison is equality, the triangles share two vertices; strict containment again contradicts equal areas. Interchanging the two triangles covers reversed inequalities.

This alignment assumes the labeled triangles have the same orientation. For mirror-image labeled triangles, include a reflection among congruences or compare them as unoriented figures. A group containing only holomorphic isometries does not realize an arbitrary reflected labeling.

Approach. Normalize the figure using an isometry, then apply the definition.

Use a real Möbius map of positive determinant to send its two ordered ideal vertices to infinity and a finite point \(a\). The remaining side through \(a\) lies on a semicircle centered at a real point \(c\), of radius \(R>0\). The affine isometry \(w\mapsto(w-c)/R\) keeps infinity fixed and sends that semicircle to the unit semicircle. Choose the endpoint order so \(a\) becomes \(-1\). The finite vertex is \(e^{i\alpha}\), and the remaining geodesic to infinity is vertical. Conformality identifies its interior angle with \(\alpha\).

Approach. Normalize the figure using an isometry, then apply the definition.

For \(T(z)=(az+b)/(cz+d)\) with real coefficients and determinant \(\Delta>0\), its real Jacobian is \(J=|T'(z)|^2=\Delta^2/|cz+d|^4\). Also \(Y=\operatorname{Im}T(z)=\Delta y/|cz+d|^2\). Thus \[\frac{dX\,dY}{Y^2}=\frac{J\,dx\,dy}{\Delta^2y^2/|cz+d|^4}=\frac{dx\,dy}{y^2}.\] The change-of-variables theorem proves equality of areas, including infinite areas by increasing compact approximations.

Approach. Normalize the figure using an isometry, then apply the definition.

The base angles are right angles and the equal summit angles are \(\alpha\), with \(0<\alpha<\pi/2\). Hence \(A=\pi-2\alpha\), so \(0<A<\pi\).

Every value occurs. Construct a symmetric Saccheri quadrilateral in Fermi coordinates around a base geodesic, with half-base \(a>0\) and leg length \(h>0\). Its summit angle satisfies \(\tan\alpha=\coth a/\sinh h\) (obtained from either half as a Lambert quadrilateral). For any chosen \(\alpha\in(0,\pi/2)\) and \(a>0\), taking \(\sinh h=\coth a/\tan\alpha\) supplies such a quadrilateral. Equivalently continuous growth of the height makes the summit angle decrease from \(\pi/2\) to \(0\).

Approach. Normalize the figure using an isometry, then apply the definition.

Normalize the common ideal point to infinity, the horocycle to \(y=h\), and the two diameters to \(x=a,b\). The sector is \(a\le x\le b\), \(y\ge h\). Its area is \[A=\int_a^b\int_h^\infty y^{-2}\,dy\,dx=\frac{b-a}{h}=d.\] The ideal cusp has finite area even though its diameters have infinite length.

Approach. Normalize the figure using an isometry, then apply the definition.

Center the circle at zero and let \(r=\tanh(R/2)\). Then \[A=\int_0^{2\pi}\int_0^r\frac{4t}{(1-t^2)^2}\,dt\,d\theta=\frac{4\pi r^2}{1-r^2}=4\pi\sinh^2(R/2).\] Since \(\sinh u>u\) for \(u>0\), this exceeds \(\pi R^2\). As \(R\to0\), \(A/(\pi R^2)\to1\).

Approach. Normalize the figure using an isometry, then apply the definition.

For a meaningful local nonempty limit, keep a point of the circles in a bounded region and let the centers recede. In the half-plane, circles with centers at hyperbolic height tending toward an ideal endpoint converge locally to a horocycle. For example circles with Euclidean center \((0,\cosh R)\) and radius \(\sinh R\), after an appropriate normalization keeping the lowest point at height \(1\), converge locally to \(y=1\).

Their geodesic curvature is \(\coth R\to1\), whereas a geodesic has curvature \(0\), so a smooth nondegenerate local limit cannot be a geodesic. Without normalization circles may simply escape all compact sets; there is no single universal set limit.

Approach. Normalize the figure using an isometry, then apply the definition.

Yes, even when “parallel” specifically means asymptotic. In the upper half-plane take the vertical geodesics \(x=0\) and \(x=1\), which share the ideal endpoint infinity. For the other two sides use the semicircles with endpoint pairs \((-1,2)\) and \((-1,3)\), which share the ideal endpoint \(-1\). Both semicircles cross both vertical lines, and their four intersections bound a nondegenerate convex geodesic quadrilateral. Each opposite pair is asymptotic. Its area is \(2\pi\) minus its four interior angles, and varies with the chosen configuration. Allowing ultraparallel opposite sides gives further examples. Rectangles are impossible because four right angles force zero area. Euclidean equivalences between equal opposite sides and parallel sides should not be assumed.

Approach. Normalize the figure using an isometry, then apply the definition.

A Euclidean square has four right angles, which is impossible here. An equiangular area-\(1\) quadrilateral must have angle \(\alpha=(2\pi-1)/4=\pi/2-1/4\). A regular quadrilateral with this angle exists: regular hyperbolic squares vary continuously from infinitesimal squares with angles approaching \(\pi/2\) to ideal squares with angles approaching zero.

Equiangular alone does not imply uniqueness. In the hyperboloid model take the four vertices \((t,\pm x,\pm y)\) with all sign combinations, \(t=\sqrt{1+x^2+y^2}\) and \(x,y>0\). Reflections in the coordinate planes make all four angles equal. For their common angle, \[\cos\alpha=\frac{xy}{\sqrt{(1+x^2)(1+y^2)}}.\] Holding this value fixed allows a continuum of unequal pairs \(x,y\), with different adjacent side lengths, so the quadrilaterals are not all congruent. Requiring regularity (\(x=y\)) restores uniqueness up to congruence.

Approach. Normalize the figure using an isometry, then apply the definition.

The area condition alone does not determine a limit; elongated or irregular polygons may behave differently. For regular convex polygons with a common center, the limit is the centered disk of area \(1\). Its radius solves \(2\pi(\cosh R-1)=1\), giving \[R=\operatorname{arcosh}\left(1+\frac1{2\pi}\right).\] To justify convergence, inscribed regular \(n\)-gons approach each fixed circle, their areas increase to the disk area, and continuity and strict monotonicity in circumradius force the area-\(1\) circumradii to this value.

Approach. Normalize the figure using an isometry, then apply the definition.

Their angle is \(\alpha=\pi/2-1/4\). An edge-to-edge tiling by these convex blocks would require an integer \(q\) with \(q\alpha=2\pi\) at each interior vertex. This would imply \(\pi=q/[2(q-4)]\), contradicting irrationality of \(\pi\). Thus they do not give such a tiling.

Use regular hyperbolic squares with angle \(2\pi/5\) instead: five meet around each vertex, yielding the regular \(\{4,5\}\) tessellation. Each square has area \(2\pi-4(2\pi/5)=2\pi/5\). More generally \(\{p,q\}\) tiles are available when \((p-2)(q-2)>4\).

Approach. Normalize the figure using an isometry, then apply the definition.

At positive distance \(d\) from a line are the two parallel lines, one on each side. From a circle of radius \(R\), normal offsets are concentric circles of radii \(R+d\) and \(R-d\) when \(0<d<R\). At \(d=R\) the inner offset collapses to the center; beyond it there is no inner distance-locus component.

Approach. Normalize the figure using an isometry, then apply the definition.

Disjoint geodesics are either asymptotic, with separation tending to zero toward their common ideal point, or ultraparallel, with a unique common perpendicular and increasing separation away from it. A constant positive normal offset from a geodesic is a hypercycle, not another geodesic. Therefore a constant-width road uses a central geodesic and two hypercycle edges.

Approach. Normalize the figure using an isometry, then apply the definition.

A circle centered at \(O\) of radius \(R\) has concentric offsets of radii \(R+d\) and, for \(d<R\), \(R-d\), because normal geodesics are its radii. For a horocycle normalize to \(y=h\). Its normals are vertical geodesics, so distance-\(d\) offsets are \(y=he^{d}\) and \(y=he^{-d}\), also horocycles with the same ideal point. Thus paired horocycles maintain constant normal separation and can model curved rails. Their arc lengths between common normals differ by an exponential factor; equal separation does not mean equal rail lengths.

Exponentiating and solving gives \(r=(e^R-1)/(e^R+1)\). The two coordinates are \(1/2\) and \(4/5\). Their transformed radius is \((4/5-1/2)/(1-(1/2)(4/5))=1/2\), so their distance is \(\log3\). The equal hyperbolic steps have Euclidean lengths 1/2 and 3/10. Visual spacing shrinks although intrinsic spacing stays constant.

Diagonals divide the polygon into n-2 ideal triangles with disjoint interiors. Each has area \(\pi\), so the total is \((n-2)\pi\). Equivalently, the polygon defect formula has n limiting angles equal to zero. Ideal vertices lie outside the plane, and every complete side has infinite intrinsic length. Length and area are different integrals; an unbounded region need not have infinite area. The finite result follows from the area integral or its limiting triangle formula, not from Euclidean screen area.