Advanced Vectors - Part 2

Let's take an analytical approach:

Let point O = $\left( 0,0 \right)$ and P = $\left( 0,r \right)$ , where $0\le r<\frac { 1 }{ 3 }$ . Since $O{ B }_{ 1 }=1$ , let's introduce an value $\alpha$ such as ${ B }_{ 1 }=\left( \cos { \alpha } , \sin { \alpha } \right)$ .

Now we can derive other points step-by-step:

$\overrightarrow { OP } =\left( r,0 \right) ,\overrightarrow { O{ B }_{ 1 } } =\left( \cos { \alpha } ,\sin { \alpha } \right) \\ \overrightarrow { P{ B }_{ 1 } } =\overrightarrow { O{ B }_{ 1 } } -\overrightarrow { OP } =\left( \cos { \alpha } -r,\sin { \alpha } \right)$

Since $\overrightarrow { A{ B }_{ 1 } } \bot \overrightarrow { A{ B }_{ 2 } } ,\overrightarrow { AP } =\overrightarrow { A{ B }_{ 1 } } +\overrightarrow { A{ B }_{ 2 } }$ , points $A,{ B }_{ 1 },{ B }_{ 2 },P$ form a rectangle. That means $\overrightarrow { P{ B }_{ 1 } } \bot \overrightarrow { P{ B }_{ 2 } }$ . Knowing that we can conclude:

$\overrightarrow { P{ B }_{ 2 } } =\left( -k\cdot \sin { \alpha } ,k\cdot \left( \cos { \alpha } -r \right) \right)$ , where $k$ is unknown real number.

To find $k$ we need to derive $\overrightarrow { O{ B }_{ 2 } }$ :

$\overrightarrow { O{ B }_{ 2 } } =\overrightarrow { OP } +\overrightarrow { P{ B }_{ 2 } } =\left( r-k\cdot \sin { \alpha } ,k\cdot \cos { \alpha } -k\cdot r \right)$

Since $\left| \overrightarrow { O{ B }_{ 2 } } \right| =1$ :

${ \left( r-k\cdot \sin { \alpha } \right) }^{ 2 }+{ \left( k\cdot \cos { \alpha } -k\cdot r \right) }^{ 2 }=1\\ { r }^{ 2 }-2\sin { \alpha } \cdot k\cdot r+{ \sin }^{ 2 }{ \alpha }\cdot { k }^{ 2 }+{ \cos }^{ 2 }{ \alpha }\cdot { k }^{ 2 }-2\cos { \alpha } \cdot { k }^{ 2 }\cdot r+{ k }^{ 2 }\cdot { r }^{ 2 }=1\\ { k }^{ 2 }-2\cos { \alpha } \cdot { k }^{ 2 }\cdot r+{ k }^{ 2 }\cdot { r }^{ 2 }-2\sin { \alpha } \cdot k\cdot r=1-{ r }^{ 2 }\\ { k }^{ 2 }\cdot \left( 1-2\cos { \alpha } \cdot r+{ r }^{ 2 } \right) -2\sin { \alpha } \cdot k\cdot r=1-{ r }^{ 2 }$

Let's stop right there and proceed with deriving the point $A$ . Since $A{ B }_{ 1 }P{ B }_{ 2 }$ is a rectangle:

$\overrightarrow { OA } =\overrightarrow { O{ B }_{ 2 } } +\overrightarrow { { B }_{ 2 }A } =\overrightarrow { O{ B }_{ 2 } } +\overrightarrow { P{ B }_{ 1 } } =\left( \cos { \alpha } -\sin { \alpha } \cdot k,\sin { \alpha } +\cos { \alpha } \cdot k-k\cdot r \right)$

Finally we can find $\left| \overrightarrow { OA } \right|$ :

${ \left| \overrightarrow { OA } \right| }^{ 2 }={ \left( \cos { \alpha } -\sin { \alpha } \cdot k \right) }^{ 2 }+{ \left( \sin { \alpha } +\cos { \alpha } \cdot k-k\cdot r \right) }^{ 2 }\\ { \left| \overrightarrow { OA } \right| }^{ 2 }={ \cos }^{ 2 }{ \alpha }-2\cos { \alpha } \sin { \alpha } \cdot k+{ \sin }^{ 2 }{ \alpha }\cdot { k }^{ 2 }+{ \sin }^{ 2 }{ \alpha }+{ \cos }^{ 2 }{ \alpha }\cdot { k }^{ 2 }+{ k }^{ 2 }\cdot { r }^{ 2 }+2\cos { \alpha } \sin { \alpha } \cdot k-2\cos { \alpha } \cdot k\cdot r-2\cos { \alpha } \cdot { k }^{ 2 }\cdot r\\ { \left| \overrightarrow { OA } \right| }^{ 2 }=1+{ k }^{ 2 }+{ k }^{ 2 }\cdot { r }^{ 2 }-2\cos { \alpha } \cdot k\cdot r-2\cos { \alpha } \cdot { k }^{ 2 }\cdot r\\ { \left| \overrightarrow { OA } \right| }^{ 2 }=1+{ k }^{ 2 }\cdot \left( 1-2\cos { \alpha } \cdot r+{ r }^{ 2 } \right) -2\sin { \alpha } \cdot k\cdot r$

Using the relation we found earlier:

${ \left| \overrightarrow { OA } \right| }^{ 2 }=1+1-{ r }^{ 2 }=2-{ r }^{ 2 }$

${ \left| \overrightarrow { OA } \right| }=\sqrt { 2-{ r }^{ 2 } }$ . Note that the distance doesn't depend on the value of $\alpha$ .

That means that:

$a=\min { { \left| \overrightarrow { OA } \right| } } =\sqrt { 2-{ \left( \frac { 1 }{ 3 } \right) }^{ 2 } } =\sqrt { \frac { 17 }{ 9 } } =\frac { \sqrt { 17 } }{ 3 } ,\quad r\rightarrow \frac { 1 }{ 3 } \\ b=\max { { \left| \overrightarrow { OA } \right| } } =\sqrt { 2-{ 0 }^{ 2 } } =\sqrt { 2 } ,\quad r=0$

So the answer is $\left\lfloor 1000\left( a+b \right) \right\rfloor = \boxed { 2788 }$