= Solution
For one triangle, equal base <spins> $s=s'=\sigma$ give <energy> $J+2\bar J\sigma\bar s$. Their minimum occurs at $\bar s=-\sigma$ and is $J-2\bar J$. Opposite base <spins> give <energy> $-J$, independently of the apex <spin>. The all-equal configurations have <energy> $J+2\bar J$.
When $\bar J=J$, the six configurations with precisely one <spin> different from the other two all have minimum <energy> $-J$. For a fixed first base <spin>, there is one minimum choice if the next base <spin> is the same, and two choices of apex if it is opposite. The ground-configuration counting <transfer matrix> is therefore
$$
C=\begin{pmatrix}1&2\\2&1\end{pmatrix},\qquad \Omega_L=\operatorname{Tr}C^L=3^L+(-1)^L.
$$
This <ground-state counting of the frustrated sawtooth Ising chain> proves the extensive <geometric frustration> and the <residual entropy> $k_B\lim_{L\to\infty}\log\Omega_L/(2L)=(k_B/2)\log3$.
When $\bar J>J$, $J-2\bar J<-J$. Every triangle must have equal base <spins> and the opposite apex <spin>. Shared base <spins> propagate that choice around the chain, leaving just two global <ground states>, of <energy> $L(J-2\bar J)$. Their <entropy> per <spin> is $k_B\log2/(2L)\to0$.
For couplings favoring <ferromagnetism>, write $J=-K$, $\bar J=-\bar K$ with $K,\bar K>0$. Both orientations of an all-equal triangle minimize every bond simultaneously. The common base <spins> then make the whole chain uniform. There are two global <ground states> for either $K=\bar K$ or $K\ne\bar K$, so no corresponding distinction in <residual entropy> occurs. This statement assumes all bonds remain nonzero.
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