13.411 Two points $\boldsymbol{A}$ and $\boldsymbol{B}$, initially at a distance of $a$ from each other, simultaneously started moving towards the vertex of a right angle along different sides with the same constant speed $v$. Point $B$ reaches the vertex $t$ units of time earlier than point $A$ (all measurements are in the same units). Determine how long point $A$ was moving. What value should be assigned to the quantity $a$ so that the sought time takes the smallest possible value?
Solution. Let $x$ be the time of movement of point $A$. Then point $A$ travels a distance $\boldsymbol{x} \cdot v$ to the vertex, and point $B-(x-t) v$.

By the Pythagorean theorem, $x^{2} v^{2}+(x-t)^{2} v^{2}=a^{2} \Rightarrow x=\frac{v t \pm \sqrt{2 a^{2}-v^{2} t^{2}}}{2 v}$.

Since $x>t$, only one value of $x$ remains, and

$\frac{v t+\sqrt{2 a^{2}-v^{2} t^{2}}}{2 v}>t$. Solving this inequality, we get: $a>v t$.

Answer: $\frac{v t+\sqrt{2 a^{2}-v^{2} t^{2}}}{2 v}$, where $a>v t$.