= Solution
Map both $a$ and $b$ to the nonidentity element of $C_2$. Both relators $a^2,b^2$ map to the identity, so this gives a homomorphism $D_\infty\to C_2$. A word of length $n$ maps to the parity class of $n$; consequently a null word has even length.
Now let $w$ be a null word of positive even length. Interpreting $a^{-1}=a$ and $b^{-1}=b$, the <normal form theorem for an amalgamated free product>[free-product normal form theorem] says that a nonempty alternating word cannot be trivial. Thus $w$ has two adjacent equal letters. Delete this $a^2$ or $b^2$, using one conjugate of a defining relator, and apply induction to the resulting null word of length $n-2$. This gives
$$
\operatorname{Area}(w)\leq1+\frac{n-2}{2}=\frac n2.
$$
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