Find the minimum perimeter of a triangle with these constraints

[songoku]

Homework Statement

Please see below

Relevant Equations

Not really sure

My attempt:
$$|\vec{BA} -\lambda \vec {BC}| \geq 2|\vec {BC}|$$
$$|\vec {BA}|^2 -2 \lambda (\vec {BA} \cdot \vec {BC}) +\lambda ^{2} |\vec {BC}|^2 \geq 4|\vec {BC}|^2$$
$$|\vec {BA}|^2 -2 \lambda |\vec {BA}| \cos \theta +\lambda ^{2} \geq 4$$

Am I even on the right track?

Thanks

 

[andrewkirk]

It's a good start
By rearranging so that the RHS is zero you get a quadratic in $\lambda$ that must always be more than zero, which means it must have no real roots, which means the discriminant ($b^2-4ac$) must be negative. That will give you some constraints on $\bar{BA}$ which you can then use to work out the minimum perimeter triangle.

 

[songoku]

It's a good start
By rearranging so that the RHS is zero you get a quadratic in $\lambda$ that must always be more than zero, which means it must have no real roots, which means the discriminant ($b^2-4ac$) must be negative. That will give you some constraints on $\bar{BA}$ which you can then use to work out the minimum perimeter triangle.

$$\lambda^2-2 \lambda |\vec {BA}| \cos \theta+|\vec {BA}|^2 -4 \geq 0$$

Using D ≤ 0 :
$$4 |\vec {BA}|^2 \cos^2 \theta-4|\vec{BA}|^2+16 \leq0$$
$$-|\vec {BA}|^2 \sin^2 \theta +4\leq0$$
$$|\vec{BA}|^2 \sin^2 \theta \geq 4$$

The maximum value of $\sin^2 \theta$ is 1 so the minimum value of $|\vec{BA}|^2$ is 4 and the minimum value of $|\vec{BA}|$ is 2

By taking $|\vec{BA}|=2$ and $\theta = \frac{\pi}{2}$, I got $|AC|=\sqrt{5}$ so the minimum perimeter is $3+\sqrt{5}$

Where is my mistake? Thanks

 

[fresh_42]

Following your strategy, I get the same (wrong) result. I guess it is the cosine that leads to the mistake. $\phi$ isn't independent of $A$ so you cannot treat it as a constant.

How about the following strategy:

We may assume $B=(0,0)$ and $C=(1,0)$ and $A=(x,y)$ with $y>0.$
Then the condition reads $\sqrt{(x-\lambda )^2+y^2}\geq 2$ which implies $y\geq 2.$
The minimal perimeter is then $1+\sqrt{x^2+y^2}+\sqrt{(x-1)^2+y^2}$ which is obviously minimal for $y=2.$ Thus we have to minimize the function $f(x)=\sqrt{x^2+4}+\sqrt{(x-1)^2+4}.$

 

[songoku]

Thank you very much andrewkirk and fresh_42