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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.

We construct a family by seven-triangle Dirichlet transplantation. Start with a scalene triangle $\tau$ sufficiently close to an equilateral triangle. Its three side lengths are the three independent parameters; restrict to an open set where its angles $\alpha ,\beta ,\gamma$ are distinct and lie between $\pi /4$ and $\pi /2$. Label the sides $a,b,c$ opposite the corresponding vertices $A,B,C$.

For each of two planar domains, begin with tile $7$ and attach the remaining congruent copies by Euclidean reflection across the indicated sides. The attachment tables areFor example, $(6,7)$ in the first $a$ row means to reflect tile $7$ across its $a$ side to obtain tile $6$. Both attachment graphs are trees: there is a central tile with three arms, each consisting of two tiles. Every side not used in the table is an exterior boundary side.

At the equilateral triangle, these seven reflected tile interiors are disjoint; at each central vertex four sectors of angle $\pi /3$ meet, leaving an exterior gap $2\pi /3$. Other boundary vertices have angles $\pi /3$ or $2\pi /3$, and distinct nonincident pieces are separated. The positive angular gaps and separations persist under sufficiently small perturbations of the three side lengths. Thus this construction really gives connected polygonal planar domains on an open three-dimensional parameter set. Their only reentrant vertices are the central vertices $A,B,C$, with interior angles $4\alpha ,4\beta ,4\gamma$; all other boundary angles are one or two triangle angles and are less than $\pi$.

Because these three reentrant angles are distinct, an isometry of the two planar domains would send each labelled central vertex to its counterpart. A planar isometry is the restriction of a Euclidean rigid motion: its differential preserves the Euclidean connection, hence is constant on the connected domain. After aligning the central triangles, such a motion fixes three noncollinear points and is the identity. But the domains are not identical. On the arm attached to side $a$, the point construction attaches the last tile across side $b$, whereas the line construction attaches it across side $c$. The former side is therefore interior in one domain and exterior in the other. This proves the two domains are nonisometric throughout this three-parameter family.

We now prove equality of the Dirichlet Laplacian spectra, including multiplicities. Index $1,\dots ,7$ by the nonzero binary vectors $x=(x_1,x_2,x_3)$, with integer label $x_1+2x_2+4x_3$. The transformationsare involutions over ${\mathbb{F}}_2$. Their point permutations $P_s$ are precisely the transpositions in the left attachment table, with the remaining points fixed. On covectors the inverse-transpose transformations give permutations $Q_s$ with exactly the right table. Let $B$ be the point-line incidence matrix of a set system of the Fano plane:The incidence relation is invariant under transforming $x$ and its covector $y$ contragrediently, so $B{P}_s=Q_{s}B$ for all three sides. Each point lies on three lines, and two distinct points lie on exactly one common line. Consequentlywhere $J$ is the all-ones matrix. Its eigenvalues are $9$ once and $2$ six times, so $B$ is invertible.

The Dirichlet boundary condition requires a sign change at an unglued side. Give each tile the sign $(-1{)}^{\text{distance from tile }7}$ and putThe signs alternate across each glued side in both domains. The signed side matrices are thereforeThey have entry $+1$ across a glued pair and entry $-1$ on a fixed exterior tile. The invertible seven-triangle Dirichlet transplantation matrix $C=DBD$ satisfiesThis is an actual Dirichlet boundary condition intertwiner, not merely the unsigned incidence relation.

To justify the spectral conclusion even at the reentrant corners, use the closed Dirichlet energy form. Pull a function on either domain back to its seven reference triangles and write its components as a vector $F$. The form domain consists of $H^1$ components whose side traces agree on glued pairs and vanish on exterior sides; equivalently $F=S_{s}F$ on each side for the point domain, and $G=T_{s}G$ for the line domain. These trace conditions exactly characterize $H_0^1$ of the glued polygon. Both its $L^2$ norm and its Dirichlet energy are the sums of their seven reference-tile integrals.

Since the signed side matrices are symmetric, the intertwining identities also give $S_s{C}^{T}C=C^{T}C{S}_s$. Thus $(C^{T}C{)}^{-1/2}$ commutes with every $S_s$, and the polar factoris an orthogonal matrix with $U{S}_s=T_{s}U$. Multiplication of the component vector by this constant matrix preserves the $L^2$ inner product and the integrated squared gradients, and maps the two form domains bijectively. The self-adjoint operators associated with these forms are consequently unitarily equivalent, so their eigenvalues and multiplicities agree. The eigenfunctions are smooth in the interiors; no unwarranted smoothness at polygonal corners is required. We have proved

Take seven copies of a scalene triangle, with sides labelled $a,b,c$ opposite its three vertices, and reflect adjacent copies across their glued side. For the first planar domain, the glued pairs are $a:(2,3),(6,7)$, $b:(4,6),(5,7)$, $c:(1,5),(3,7)$; for the second they are $a:(1,3),(5,7)$, $b:(2,6),(3,7)$, $c:(4,5),(6,7)$. Both coloured trees have central tile $7$. For triangles sufficiently close to equilateral the interiors do not overlap. If its three angles are distinct and between $\pi /4$ and $\pi /2$, the only reentrant vertices are the original central vertices, with angles four times the corresponding triangle angles. They force any proposed isometry to preserve the labelled central triangle, which is incompatible with the different attachments on its arms.

For the spectral proof index the nonzero vectors and covectors of ${\mathbb{F}}_2^3$ by $1,\dots ,7$, using binary coordinates. The Fano plane incidence matrix is $B_{y,x}=1_{y\cdot x=0}$. It satisfies $B^{T}B=2I+J$, so is invertible, and intertwines the point and line permutations of each of the three elementary transvections. Put $D=\mathrm{diag}(1,1,-1,1,-1,-1,1)$. The signed reflection matrices for the Dirichlet boundary condition are $-D{P}_{s}D$ and $-D{Q}_{s}D$, hence $C=DBD$ intertwines them. Its polar orthogonal factor gives a unitary equivalence of the Dirichlet Laplacians, preserving both the $L^2$ inner product and the Dirichlet energy of the piecewise functions. This produces a three-parameter family of noncongruent isospectral planar domains.