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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.

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.