Let $O_1$ , $O_2$ , $O_3$ be a circle of radius $1$.
Each of $O_1$ , $O_2$ , $O_3$ meet the others. (Please see figure)
Let
P : the point of circle $O_2$
Q : the point of circle $O_3$
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In this case,
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What is the maximum of $\vec{O_1 P} \cdot \vec{O_1 Q} $ (inner product)
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I've struggled this for hours
and I get enter link description here
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But this is for high school students not for university ...
How do I deal with this....?
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Thanks in advance.
It's not trivial, but it can be done.
Step one: let's fix the position of $Q$, and for now choose $O_1Q$ as the $x$ axis of the coordinate system, such that $O_1$ is at $0$. The dot product that you want to maximize in this case is $O_1Q$ (fixed) multiplied with the $x$ coordinate of the $P$ point. You can now calculate the $x$ position of the $P$ point, as the $x$ coordinate of the $O_2$ plus $1\cdot\cos\theta_P$. The first part does not depend on $\theta_P$, so the maximum is achieved when $\theta_P =0$, or in geometric terms $O_2P||O_1Q$.
Step two: repeat the same procedure by fixing $P$, to obtain $O_3Q||O_1P$.
Step three: since the above conditions are independent of each other, the maximum is achieved when both are true. Let's calculate the coordinate of $P$ and $Q$. For simplicity, use the coordinate system with the $x$ axis starting at $O_1$ and perpendicular to $O_2O_3$. If you extend $O_2P$ and $O_3Q$, they intersect the $x$ axis at the same point $O_4$, forming a parallelogram $O_1O_2O_4O_3$, with $O_1O_4$ as one of the diagonals. The other diagonal intersects the $x$ axis at point $M$. So you have $O_1O_2=2$, $O_2M=1$, therefore $O_1M=\sqrt{3}$. By symmetry, the $x$ coordinates of $P$ and $Q$ are $\frac{3}{2}O_1M=\frac{3}{2}\sqrt{3}$. Also by symmetry, you can get the $y$ coordinates as $\pm\frac{1}{2}$.
Step four: $$\vec{O_1P}\cdot\vec{O_1Q}=x_Px_Q+y_Py_Q=\frac{\sqrt{27}}{2}\frac{\sqrt{27}}{2}+\frac{1}{2}\frac{-1}{2}=\frac{26}{4}$$
That's if I did not make any errors :)