This solution assumes a basic understanding of vectors.

In this problem, Takahashi and Aoki always move in a straight line at constant speed. For linear motion, if a point starts at position \(\bm{s_0}\) and moves with constant velocity \(\bm{v}\), then its position at time \(t\geq0\) is \(\bm{s}(t)=\bm{s_0}+t\bm{v}.\)

We will view start and goal points \(\text{TS}, \text{TG}, \text{AS}, \text{AG}\) as two-dimensional vectors. Define \(t_\text{T}=|\text{TG}-\text{TS}|\) as the total time Takahashi spends walking, and define \(\bm{v}_\text{T}=\frac1{t_\text{T}}(\text{TG}-\text{TS})\) as Takahashi’s velocity. Similarly define \(t_\text{A}\) and \(\bm{v}_\text{A}\) as Aoki’s total time walking and Aoki’s velocity, respectively.

Since Takahashi walks in a straight line, his position vector \(\bm{s}_\text{T}(t)\) at time \(t\in[0, t_\text{T}]\) is given by the vector equation

\[\bm{s}_\text{T}(t)=\text{TS}+t\bm{v}_\text{T}.\]

Similarly, Aoki’s position vector \(\bm{s}_\text{A}(t)\) at time \(t\in[0, t_\text{A}]\) is given by

\[\bm{s}_\text{A}(t)=\text{AS}+t\bm{v}_\text{A}.\]

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By symmetry, if \(t_\text{T}\gt t_\text{A}\), we can swap Takahashi and Aoki’s roles and assume \(t_\text{T}\leq t_\text{A}\).

As the problem suggests, Takahashi and Aoki do not necessarily stop at the same time. We can divide the problem into two time intervals:

• For \(t\in[0, t_\text{T}]\), both Takahashi and Aoki are walking towards their goal points.
• For \(t\in(t_\text{T}, t_\text{A}]\), Takahashi is stationary and Aoki is walking towards his goal point.

If \(t_\text{T}=t_\text{A}\), the second interval is empty, so the answer is the minimum distance in the first time interval. Otherwise, the answer is minimum distance among the two time intervals.

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Consider the first time interval \(t\in[0, t_\text{T}]\). From Takahashi’s perspective, Aoki’s relative starting point is \((\text{AS}-\text{TS})\), and Aoki’s relative velocity is \((\bm{v}_\text{A}-\bm{v}_\text{T})\). Thus, Aoki’s relative position is

\[\bm{s}_\text{A}(t)-\bm{s}_\text{T}(t)=(\text{AS}-\text{TS})+t(\bm{v}_\text{A}-\bm{v}_\text{T}).\]

Since this equation describes linear motion, we can model Aoki’s relative movement as a line segment for \(t\in[0, t_\text{T}]\).

Next, consider the second time interval \(t\in(t_\text{T}, t_\text{A}]\). Since Takahashi stays at point \(\text{TG}\), we can model Aoki’s motion from the perspective of stationary point \(\text{TG}\). Aoki’s relative starting point is \((\text{AS}-\text{TG})\) and his relative velocity is \(\bm{v}_\text{A}\), so Aoki’s relative position is

\[\bm{s}_\text{A}(t)-\text{TG}=(\text{AS}-\text{TG})+t\bm{v}_\text{A}.\]

Since this equation similarly describes linear motion, we can again model Aoki’s relative movement as a line segment for \(t\in(t_\text{T}, t_\text{A}]\).

In both time intervals, we can rephrase the problem as finding the minimum distance between the origin and a given line segment.

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Consider a line segment with distinct endpoints \(\bm{p}=(x_p, y_p)\) and \(\bm{q}=(x_q, y_q)\). Our goal is to find the minimum distance between the origin and the given line segment.

We can represent the line segment as a vector equation \(L(\lambda)=\bm{p}+\lambda(\bm{q}-\bm{p})\), giving us a point on the line segment for parameter \(\lambda\in[0, 1]\).

The distance between the origin and a point \(L(\lambda)\) is

\[|L(\lambda)|=\sqrt{(x_p+\lambda(x_q-x_p))^2+(y_p+\lambda(y_q-y_p))^2}.\]

To minimize \(|L(\lambda)|\), we can solve for \(\frac{\text{d}}{\text{d}\lambda}|L(\lambda)|=0\), but the derivative is slightly complicated. We instead utilize the fact that \(|L(\lambda)|\) is minimum whenever \(|L(\lambda)|^2\) is minimum, and use this to solve for \(\frac{\text{d}}{\text{d}\lambda}|L(\lambda)|^2=0\). Computing the derivative yields

\[ \begin{aligned} \frac{\text{d}}{\text{d}\lambda}|L(\lambda)|^2&=\frac{\text{d}}{\text{d}\lambda}\left((x_p+\lambda(x_q-x_p))^2+(y_p+\lambda(y_q-y_p))^2\right)\\ &=2(x_q-x_p)(x_p+\lambda(x_q-x_p))+2(y_q-y_p)(y_p+\lambda(y_q-y_p))\\ &=0 \end{aligned} \]

After simplifying, we obtain

\[\lambda=-\frac{x_p(x_q-x_p)+y_p(y_q-y_p)}{(x_q-x_p)^2+(y_q-y_p)^2}=-\frac{\bm{p}\cdot(\bm{q}-\bm{p})}{|\bm{q}-\bm{p}|^2}=\frac{\bm{p}\cdot(\bm{p}-\bm{q})}{|\bm{p}-\bm{q}|^2}\]

that minimizes \(|L(\lambda)|\).

Note that the above \(\lambda\) may not fall within interval \([0, 1]\). If \(\lambda<0\), the closest point to the origin is \(L(0)=\bm{p}\). If \(\lambda>1\), the closest point is \(L(1)=\bm{q}\).

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Thus, we obtain a solution to the original problem. The time complexity is \(O(1)\).

Sample code (Python)