Beginner Vector Geometry #6

This problem is solvable by intuition, which is why I put this problem as "beginner".

However, the solution will be as logical as possible.

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Note that $P$ is a point on $C_2$ .

And it would be nice to have some "constant" value.

Then know that $\overrightarrow{O_1P}=\overrightarrow{O_2P}-\overrightarrow{O_2O_1}$ .

Therefore, $\overrightarrow{O_1P}\cdot\overrightarrow{O_2P}=\left(\overrightarrow{O_2P}-\overrightarrow{O_2O_1}\right)\cdot\overrightarrow{O_2P}=\overrightarrow{O_2P}^2-\overrightarrow{O_2O_1}\cdot\overrightarrow{O_2P}=25-\overrightarrow{O_2O_1}\cdot\overrightarrow{O_2P}$ .

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You'll see that $\overrightarrow{O_2O_1}\cdot\overrightarrow{O_2P}$ should be minimum.

Let the angle that $\overrightarrow{O_2O_1}$ and $\overrightarrow{O_2P}$ form be $\theta$ . $(0<\theta<\pi)$

$\overrightarrow{O_2O_1}\cdot\overrightarrow{O_2P}=|\overrightarrow{O_2O_1}||\overrightarrow{O_2P}|\cos\theta=40\cos\theta$ .

The value is minimum if $\cos\theta=-1$ .

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Therefore, the maximum of $\overrightarrow{O_1P}\cdot\overrightarrow{O_2P}$ is $25-40\times(-1)=\boxed{65}$ .