Determine whethere the value of $(\overrightarrow{BM}+\overrightarrow{BN})\bullet\overrightarrow{BP}$ is fixed

Question

Consider the circle with equation $(x+1)^2+(y-2)^2=20$, and let $A$ denotes the center of the circle as shown. Let $l_1$ be a line tangent to the circle whose equation is $x+2y+7=0$. Fix a point $B=(-2,0)$, let $l$ be any line passing through $B$, intersecting the circle at $M$ and $N$, and intersecting $l_1$ at $P$. Determine whethere the value of $(\overrightarrow{BM}+\overrightarrow{BN})\bullet\overrightarrow{BP}$ is fixed, i.e. independent of the choince of $l$.

In the picture, $Q$ denotes the mid point of the line segment $\overline{MN}$.

First I apologize that the picture is not that clear, this question is from my collegue and I don't have the original resourse.

About my thinking: The very first thing I did is plug in some specific choices of $M,N,P$ and calculated the correspoding values. For example I chose $l$ to be the line parallel to the $y-$axis and then chose it to be the line parallel to the $x-$axis (which is exactly the $x-$axis). In both cases I obtain $(\overrightarrow{BM}+\overrightarrow{BN})\bullet\overrightarrow{BP}=-10$. Therefore I guess that the answer is positive, i.e. $(\overrightarrow{BM}+\overrightarrow{BN})\bullet\overrightarrow{BP}=-10$, no matter how we choose the line $l$. But I cannot prove or disprove the conjecture. Any hints to this problem? Thanks in advance.

Answer 1

$$(\overrightarrow{BM}+\overrightarrow{BN})\bullet\overrightarrow{BP}=\overrightarrow{BM}\bullet\overrightarrow{BP}+\overrightarrow{BN}\bullet\overrightarrow{BP}$$

And using that $\vec{u}\bullet \vec{v}=|\vec{u}|\cdot|\vec{v}|\cos \alpha$ we get:

$$(\overrightarrow{BM}+\overrightarrow{BN})\bullet\overrightarrow{BP}=-BM\cdot BP +BN\cdot BP=BP\cdot(BN-BM)$$

Now using the power of the point $P$ we have:

$$PN\cdot PM=PT^2 \rightarrow (BP-BM)\cdot(BP+BM)=PT^2$$ $$BP(BN-BM)=BP^2-PT^2-BN\cdot BM \quad (1)$$

But $BN\cdot BM$ is the power of the point $B$ and it is equal to $R^2-AB^2$.

Futhermore if we look to the line $\overleftrightarrow{AB}$ its angular coeficient is $2$. That means the lines $\overleftrightarrow{AB}$ and $l_1$ are perpendicular and then from Pythagoras theorem at the triangule $PBT$ we get:

$$BP^2=PT^2+AB^2 \rightarrow BP^2-PT^2=AB^2$$

And backing to $(1)$ we have:

$$BP(BN-BM)=AB^2-(R^2-AB^2)=2\cdot AB^2-R^2=2\cdot 5-20=-10$$

In fact, the value of $(\overrightarrow{BM}+\overrightarrow{BN})\bullet\overrightarrow{BP}$ doesn't depend on the choice of $l$.

Answer 2

Let $E$ be the midpoint of segment $MN$ and let circle and tangent touch at $D$. Then $$\overrightarrow{BM}+\overrightarrow{BN}={1\over 2}\overrightarrow{BE}$$

Now since $$\angle PEA = \angle PDA = 90^{\circ}\;\;\;\;\Longrightarrow \;\;\;\; ACDE {\rm \;is\; cyclic}$$

So by the power of the point $B$ with respect to the circle $ACDE$ we have $$ \overrightarrow{BE} \cdot \overrightarrow{BP} = \overrightarrow{BD}\cdot \overrightarrow{BA} = constant$$