Problem 1807

The answer is $t = 30 + 30\sqrt{5}$.

We work in the Cartesian plane, and we let $\mathcal{B}$ denote the boundary of the unit square with corners $(\pm \frac{1}{2}, \pm \frac{1}{2})$. We complete the solution in three steps.

Step 1:
Let $A_1 = (-\frac{1}{2}, 0)$ and $A_2 = (\frac{1}{2}, 0)$, and consider the function $f(P) = |A_1P| + |A_2P|$ for $P \in \mathcal{B}$. It is not difficult to see that by symmetry we may assume that $f(P)$ attains its maximum in the right half of the upper side of $\mathcal{B}$, in some point $P = (p, \frac{1}{2})$ with $0 \le p \le \frac{1}{2}$. Then, we have
$$f(P) = \sqrt{\left(\frac{1}{2} - p\right)^2 + \frac{1}{4}} + \sqrt{\left(\frac{1}{2} + p\right)^2 + \frac{1}{4}}.$$
The function $f(P)$ is non-negative, and its square $1 + 2p^2 + \sqrt{p^4 + \frac{1}{4}}$ is increasing in $p$ for $p \ge 0$. Hence $f(P)$ is increasing in $p$ for $0 \le p \le \frac{1}{2}$, and $f(P) \le f(\frac{1}{2}, \frac{1}{2})$ holds for all $P \in \mathcal{B}$. Applying this repeatedly to the points $P_1, \dots, P_{120}$, we see that
$$\sum_{k=1}^{120} |A_1 P_k| + |A_2 P_k| \le 120 \cdot f\left(\frac{1}{2}, \frac{1}{2}\right) = 60 + 60\sqrt{5}.$$
Consequently for some $Q \in \{A_1, A_2\}$, we have $\sum_{k=1}^{120} |QP_k| \le 30 + 30\sqrt{5}$.

Step 2:
Let $B_1, B_2, B_3, B_4$ be the corners of the square, and let $g(P) = |B_1P| + |B_2P| + |B_3P| + |B_4P|$ for $P \in \mathcal{B}$. By symmetry considerations, $g(P)$ attains its minimum in the right half of the upper side of $\mathcal{B}$. We may therefore assume that this minimum occurs at some point $P = (p, \frac{1}{2})$ with $0 \le p \le \frac{1}{2}$. Then, we have
$$g(P) = 1 + \sqrt{\left(\frac{1}{2} - p\right)^2 + 1} + \sqrt{\left(\frac{1}{2} + p\right)^2 + 1}.$$
As in Step 1, we see that $g(P)$ is increasing in $p$ for $0 \le p \le \frac{1}{2}$. This yields $g(P) \ge g(0, \frac{1}{2})$ for all $P \in \mathcal{B}$, and hence
$$\sum_{k=1}^{120} |B_1 P_k| + |B_2 P_k| + |B_3 P_k| + |B_4 P_k| \ge 120 \cdot g\left(0, \frac{1}{2}\right) = 120 + 120\sqrt{5}.$$
Consequently for some $Q \in \{B_1, B_2, B_3, B_4\}$, we have $\sum_{k=1}^{120} |QP_k| \ge 30 + 30\sqrt{5}$.

Step 3:
Finally, we argue that $t = 30 + 30\sqrt{5}$ satisfies the property in the problem statement. For points $Q$ lying on the square, the value of $|P_1Q| + \dots + |P_{120}Q|$ is a continuous function of $Q$. By Step 1, there exists some $Q_1$ for which this value is at most $t$, and by Step 2, there exists some $Q_2$ for which this value is at least $t$. Thus, by the intermediate value theorem, if we move $Q$ from $Q_1$ to $Q_2$, there exists an intermediate point $Q$ for which this value is exactly $t$, as needed.