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Put $q=|z{|}^2$ and retain the domain restriction $z\ne 0$. Then $z+1/\bar{z}=(q+1)/\bar{z}$, so the continued fraction simplifies toThe other denominators are nonzero automatically because $q+1,q+2>0$. Writing $z=x+iy$, the real-part condition becomesMultiplying by $q(q+2)$ and rearranging gives $q+4x+2=0$. Thus the first geometric locus in the complex plane is the circleIts center is $-2$ and radius is $\sqrt{2}$; it does not contain the excluded point $z=0$ on its circumference.

The principal complex logarithm has imaginary part equal to the complex argument, so its equation gives the open rayIf one interprets the logarithm as multivalued, requiring a value with imaginary part $\pi /3$ gives the same ray. Expanding $z^2$ in the other equation gives $2ixy=2ix$, hence $x(y-1)=0$. Its solution is the union of the entire imaginary axis and the horizontal line $y=1$.

These latter loci bound a right triangle with verticesThe bounded open region is $0<x<1/\sqrt{3}$, $\sqrt{3}x<y<1$. Its closure is the closed triangle. Although the logarithm is undefined at $O$, the imaginary-axis locus includes that vertex. The earlier circle is a separate locus and is not a side of this triangle.

The right angle is at $P$, so the circumcenter is the midpoint of hypotenuse $OQ$:Thus its circumcircle has the complex equationTo find the incircle, let its radius be $r$. Tangency to $x=0$ and $y=1$ puts its center at $z_i=r+i(1-r)$. The distance from that point to $y-\sqrt{3}x=0$ is $(1-r-\sqrt{3}r)/2$. Requiring this distance to be $r$ givesThereforeThis tangent circle is indeed the largest contained circle. For any contained disk of radius $\rho$, its center has distances $d_j\ge \rho$ from the three side lines. Splitting the triangle into triangles with that center as vertex gives area $\Delta =\frac{1}{2}{\sum }_j{\ell }_j{d}_j\ge s\rho$, where $s$ is its semiperimeter. Here $\Delta =1/(2\sqrt{3})$ and $s=(1+\sqrt{3})/2$, so $\rho \le \Delta /s=r$. Our circle attains this upper bound, proving maximality rather than just tangency.

We construct the two domains explicitly from seven congruent triangles. Let a scalene reference triangle have sides $a,b,c$ opposite its three labelled vertices. Choose its three side lengths in a small open neighbourhood of $(1,1,1)$, with a fixed strict ordering. Thus the three angles are distinct and lie between $\pi /4$ and $\pi /2$. Each copy keeps these side labels.

For the first assembly glue the following tile pairs, reflecting across the indicated side:Each coloured adjacency graph is a tree with central tile $7$ and three two-tile arms. Place tile $7$ in the plane and obtain every other tile by the prescribed reflections. At the equilateral reference shape the copies have disjoint interiors and form a polygonal disk; the same holds throughout a sufficiently small neighbourhood, since the only contacts between tiles are the prescribed edges and their incident vertices. In particular each assembly is a bounded connected planar domain.

To prove equality of spectra, index tiles by the seven nonzero vectors $x\in {\mathbb{F}}_2^3$, using $x=(x_1,x_2,x_3)$ with label $x_1+2x_2+4x_3$. LetThese involutions act on points by $x\mapsto s_{s}x$ and on nonzero covectors by $y\mapsto s_s^{-T}y=s_s^{T}y$. Their swaps are exactly the two columns of the table. Write $P_s,Q_s$ for the corresponding permutation matrices. The Fano plane incidence matrixsatisfies $B{P}_s=Q_{s}B$, since incidence is preserved by a linear transformation and its inverse transpose. Each point lies on three lines, while two different points have one common line. Thereforewhere $J$ is the all-ones matrix, proving that $B$ is invertible.

For the Dirichlet boundary condition, a glued side has a positive swap and an exposed side has diagonal entry $-1$. Both trees have the same bipartition, with sign matrixTheir signed reflection matrices are $M_s=-D{P}_{s}D$ and $N_s=-D{Q}_{s}D$. On a glued edge its endpoint signs are opposite, so the off-diagonal entries are $+1$; on an exposed edge the entry is $-1$. Consequently the invertible matrix $C=DBD$ satisfies

We spell out why this gives isospectrality. Pull each function back from the seven tiles to the reference triangle. Its seven boundary values along side $s$ form a vector $u_s$ satisfying $M_s{u}_s=u_s$: this imposes equal values at glued edges and zero values at exposed edges. Outward normal derivatives of a smooth eigenfunction satisfy $M_s{v}_s=-v_s$, giving opposite normal derivatives on glued edges. The identities above preserve both conditions under $u\mapsto Cu$. Since $C$ is constant, it also commutes with the scalar Laplacian on each triangle.

One can include all weak eigenfunctions without any corner regularity issue by orthogonalization of a transplantation matrix. The matrix $C^{T}C$ commutes with the symmetric $M_s$, so $O=C(C^{T}C{)}^{-1/2}$ is orthogonal and still intertwines $M_s,N_s$. On the disjoint tiles, $O$ preserves the $L^2$ norm and the sum of the Dirichlet energies; it maps the trace-matching form domain $H_0^1({\Omega }_P)$ bijectively to $H_0^1({\Omega }_L)$. Thus it is a unitary equivalence of the Dirichlet Laplacians, proving equality of every eigenvalue with its multiplicity.

Finally, the only reentrant vertices of either polygon are the three vertices of tile $7$, with interior angles four times the corresponding reference-triangle angles. All other boundary angles are less than $\pi$. These three distinct angles force any hypothetical isometry to map each labelled central vertex to its counterpart. An isometry between connected open planar domains is a restriction of a rigid motion: its derivative is orthogonal, and preservation of the flat Levi-Civita connection makes that derivative constant. After aligning the two central triangles, this rigid motion fixes three noncollinear points and is the identity. But the two domains do not coincide: the tile beyond side $a$ of the central triangle has its next attachment along $b$ in the point assembly and along $c$ in the line assembly. These occupy different open regions. This contradiction proves nonisometry.

Therefore the three freely varying side lengths give a three-parameter family of nonisometric Dirichlet-isospectral bounded planar domains. The parameters include scale; no congruence normalization removes it.

Store a bounding volume hierarchy for each object, using axis-aligned bounding boxes or tighter enclosing volumes. A leaf contains one triangle or a small group; every internal node encloses all its descendants. Start with the two roots. If their bounding volumes are disjoint, discard that pair. Otherwise descend into children of one or both nodes, usually splitting the larger node, until overlapping leaf pairs are reached. Apply the exact triangle-triangle intersection algorithm only at these leaves, and stop at the first hit if only a yes/no answer is required. The pruning is sound because disjoint enclosing volumes imply disjoint enclosed triangles.

A balanced bounding volume hierarchy typically takes $O(n\log n)$ work to construct and $O(n)$ storage. It can be reused for repeated queries and refitted when vertex positions move; a rigid transformation can also be handled in a common coordinate frame. Spatial grids or octrees provide another broad-phase strategy: place triangles in every cell they meet and test only pairs from common cells, suppressing duplicates. A box test is especially valuable because the small randomly distributed triangles almost always fail it.

For $n,m$ of order $10^5$, the naive method has about $10^{10}$ candidate pairs. Hierarchical pruning replaces this by the number of overlapping node pairs and actual candidate leaf pairs. No universal subquadratic worst-case bound follows: interpenetrating or very poorly separated meshes can still generate $nm$ candidates. If “intersection of objects” means overlap of solid volumes rather than intersection of their boundaries, add a point-in-solid containment test when no surface intersections are found; one closed object can lie wholly inside the other. Use spatial enclosure to prune pairs before exact triangle tests.

Count scalar additions, subtractions, multiplications and divisions for the specific triangle-triangle intersection algorithm above; comparisons and index selection are separate. Computing one plane normal vector uses two vector differences, costing six subtractions, and one cross product, costing six multiplications and three subtractions. The three vertex-to-plane evaluations use nine further coordinate subtractions and three dot products, each with three multiplications and two additions. The first rejection test therefore costsFor randomly located triangles of characteristic size $a$ in a box of size $H\gg a$, a second triangle straddles the first triangle's plane only when its position is in a slab of relative thickness $O(a/H)$. Consequently the first plane test rejects almost every pair: the expected arithmetic count for this implementation is $39+O(a/H)$, with the constant in the remainder determined by the subsequent branches. This is an average count, rather than the assertion that every input needs 39 operations.

For a generic surviving noncoplanar pair, the symmetric plane test costs another 39 operations, the cross product $N_A\times N_B$ costs nine, and each of the four slice endpoints needs five scalar operations in the selected coordinate: one subtraction and division for the interpolation fraction, then one subtraction, multiplication and addition for the coordinate. Thus this branch usesA fully executed coplanar test can use six planar edge directions. Each needs two subtractions to form its perpendicular and six two-dimensional dot products, costing $2+6(3)=20$ operations. Including both plane tests and the parallelism cross product gives an upper count $78+9+120=207$ before any early exit. Contact and degeneracy can shorten or change these counts. Precomputed normal vectors or a preliminary axis-aligned bounding box test also change the answer: the latter uses only comparisons after its coordinate extrema are available. The relevant small-triangle average is 39 operations for the specified plane-first implementation; a generic complete test uses 107.

Let the triangles have vertices $A_0,A_1,A_2$ and $B_0,B_1,B_2$. Work initially with nonzero-area triangles. A triangle-triangle intersection algorithm begins with the unnormalized normal vector $N_A=(A_1-A_0)\times (A_2-A_0)$ and the three signed plane evaluations $d_j=N_A\cdot (B_j-A_0)$. If all $d_j$ are strictly positive, or all strictly negative, the second triangle lies strictly on one side of the first one's plane, so the triangles are disjoint. Otherwise compute $N_B$ and $e_i=N_B\cdot (A_i-B_0)$ and apply the symmetric rejection test. Strict inequalities are important: contact on a vertex or edge counts as intersection.

If the planes are not parallel, put $L=N_A\times N_B$. Each triangle meets the other one's plane in an empty set, a point or a line segment. Its endpoints are obtained from zero-valued vertices and the edges whose endpoint plane evaluations have opposite signs. For example, an edge $B_j{B}_k$ contributesAll these points lie on the common line of the two planes. Choose a coordinate $r$ for which $L_r\ne 0$ and form the coordinate intervals $I_A=[a_{-},a_{+}]$, $I_B=[b_{-},b_{+}]$ of the two slices. This coordinate is one-to-one along the common line, so the exact final test isOnly that coordinate of each slice endpoint needs to be evaluated; no full three-dimensional line parameterization is necessary.

Parallel planes which have survived the strict plane tests are coplanar. Drop the coordinate corresponding to the largest component of $N_A$ and test the two projected planar triangles. For each edge of either projected triangle, project both vertex sets onto the perpendicular direction. A strictly disjoint pair of projection intervals certifies separation. If none of the six edge directions separates them, the planar convex sets overlap: a separating line for two disjoint convex polygons can be chosen parallel to an edge of one of them. This includes containment and boundary contact.

A zero normal vector signifies a degenerate triangle; reduce its convex hull to a point or its extreme line segment and use point containment or segment intersection instead. For floating-point input, reliable sign predicates and scale-aware error bounds are preferable to treating a tiny signed distance as an arbitrary exact zero. Plane rejection followed by interval overlap, with a planar test for coplanar input, determines all nondegenerate cases.