
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Let $C$ be the point on the highway where Kolya got off his bike, with $AC=10t$ and $D$ be the point where Vasya was at time $t$, with $AD=5t$. Let $E$ be the point where Petya picked up Vasya on the bike.

On the segment $DC$, with length $|DC|=10t-5t=5t$, Vasya and Petya approached each other at a speed of 20 km/h, and they met at point $E$ at time $t + \frac{10t - 5t}{20} = \frac{5t}{4}$.

Given that $DE = 5 \cdot \frac{t}{4}$, $EC = 15 \cdot \frac{t}{4}$, $CB = 20 - 10t$, and $BE = \frac{15t}{4} + 20 - 10t = 20 - \frac{25t}{4}$,

Vasya and Petya's trip end time is $T_{1} = t + \frac{t}{4} + \frac{20 - \frac{25t}{4}}{10} = 2 + \frac{5t}{8}$.

Kolya's trip end time is $T_{2} = t + \frac{20 - 10t}{5} = 4 - t$

Thus, $T = T(t) = \max\left(T_{1}, T_{2}\right) = \begin{cases}
T_{1} = 2 + \frac{5t}{8}, & t \geq \frac{16}{13} \\
T_{2} = 4 - t, & 0 \leq t < \frac{16}{13}
\end{cases}$

Present the answer in LaTex format: \boxed{Your answer}
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