= Solution
At $138\,{}^\circ\mathrm C$, one line is tangent to all three <Gibbs free energy> curves at approximately $x_{\rm Sn}=0.03$ for (Bi), $0.61$ for <liquid>, and $0.93$ for $(\beta\mathrm{Sn})$. These three phases can coexist at the <eutectic point>. Between the two <solid> compositions, the equilibrium phase fractions depend on how much <liquid> has already transformed; exactly at the invariant <temperature>, the overall composition alone does not fix all three fractions. At the eutectic overall composition, cooling just below the <temperature> eliminates the <liquid> and leaves the two solids.
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