Stopping Brownian motion without anticipation as close as possible to its ultimate maximum

Let $B=(B_t)_{0 \le t \le 1}$ be the standard Brownian motion started at 0, and let $S_t=\max_{ 0 \le r \le t} B_r$ for $0 \le t \le 1$. Consider the optimal stopping problem
$$ V_*=\inf_\tau{\mathsf E}(B_\tau-S_1)^2, $$
where the infimum is taken over all stopping times of $B$ satisfying $0 \le \tau \le 1$. We show that the infimum is attained at the stopping time $\tau_*=\inf\{0\le t\le 1\mid S_t-B_t\ge z_*\sqrt{1-t}\}$, where $z_*=1.12 \ldots$ is a unique root of the equation $4\Phi(z_*)-2z_*\varphi(z_*)-3=0$ with $\varphi(x)=(1/\sqrt{2 \pi })\,e^{-x^2/2}$ and $ \Phi (x)=\int_{-\infty}^x \varphi(y) dy$. The value $V_*$ equals $2 \Phi (z_*)-1$. The method of proof relies upon a stochastic integral representation of $S_1$, time-change arguments, and the solution of a free-boundary (Stefan) problem.