China Mathematical Competition (Complementary Test)

We may assume that $|P_0{P}_1|\le |P_0{P}_2|\le \cdots \le |P_0{P}_n|$. At first, we will prove that $|P_0{P}_k|>\frac{d}{3}\sqrt{k+1}$ for any positive integer $n$. Obviously, $|P_0{P}_k|\ge d\ge \frac{d}{3}\sqrt{k+1}$ for $k=1,2,\dots ,8$, and the second equality holds only when $k=8$. Then we only need to prove that $|P_0{P}_k|\ge d\ge \frac{d}{3}\sqrt{k+1}$ for $k\ge 9$. Take each $P_i$ ($i=0,1,2,\dots ,k$) as the center to draw a circle with radius $\frac{d}{2}$. Then these circles are either externally tangent to or apart from each other. Take $P_0$ as the center to draw a circle with radius $|P_0{P}_k|+\frac{d}{2}$. Then the previous $k+1$ smaller circles are all located in this larger one. Then $\pi (|P_0{P}_k|+\frac{d}{2}{)}^2>(k+1)\pi (\frac{d}{2}{)}^2$, from which we have $|P_0{P}_k|>\frac{d}{2}(\sqrt{k+1}-1)$. It is easy to check that $\frac{\sqrt{k+1}-1}{2}>\frac{\sqrt{k+1}}{3}$ for $k\ge 9$. Then $|P_0{P}_k|>\frac{d}{3}\sqrt{k+1}$ for $k\ge 9$. Over all, we have $|P_0{P}_k|>\frac{d}{3}\sqrt{k+1}$ for $k\ge 9$. Therefore, $$|P_0{P}_1|\cdot |P_0{P}_2|\cdots |P_0{P}_n|>{\left(\frac{d}{3}\right)}^n\sqrt{(n+1)!}$$

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Alternative solution.

We may assume $|P_0{P}_1|\le |P_0{P}_2|\le \cdots \le |P_0{P}_n|$. Take each $P_i$ ($i=0,1,2,\dots ,k$) as the center to draw a circle with radius $\frac{d}{2}$. Then these circles are either externally tangent to or apart from each other. Let $Q$ be any point on $\odot P_i$. Since $$\begin{array}{rl}|P_{0}Q|& \le |P_0{P}_i|+|P_{i}Q|=|P_0{P}_i|+\frac{d}{2}\\ & \le |P_0{P}_k|+\frac{1}{2}|P_0{P}_k|=\frac{3}{2}|P_0{P}_k|,\end{array}$$ we get that the circle with center $P_0$ and radius $\frac{3}{2}|P_0{P}_k|$ cover the previous $k+1$ smaller circles. Then we have $\pi (\frac{3}{2}|P_0{P}_k|{)}^2>(k+1)\pi (\frac{d}{2}{)}^2$, and that is $$|P_0{P}_k|>\frac{d}{3}\sqrt{k+1}(i=0,1,2,\dots ,k).$$ Therefore, $$|P_0{P}_1|\cdot |P_0{P}_2|\cdots |P_0{P}_n|>{\left(\frac{d}{3}\right)}^n\sqrt{(n+1)!}$$