Find the minimum perimeter of a triangle with these constraints

• songoku
I found a different solution which minimizes ##|\bar{BA}|##. First, we find the norm of ##\vec{BA}##: $$|\vec{BA}|^2=\vec{B}^2+\vec{A}^2$$ Next, set ##\vec{BA}=\vec{B}-\vec{A}## and solve for ##\vec{BA}##: $$|\vec{BA}|=\vec{B}-\vec{A}\cdot \vec{BC}-\lambda \vec {BC}=\vec{BC}-\lambda ^{2}=\vec{Af songoku Homework Statement Please see below Relevant Equations Not really sure My attempt:$$|\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$$Am I even on the right track? Thanks It's a good start By rearranging so that the RHS is zero you get a quadratic in ##\lambda## that must always be more than zero, which means it must have no real roots, which means the discriminant (##b^2-4ac##) must be negative. That will give you some constraints on ##\bar{BA}## which you can then use to work out the minimum perimeter triangle. songoku It's a good start By rearranging so that the RHS is zero you get a quadratic in ##\lambda## that must always be more than zero, which means it must have no real roots, which means the discriminant (##b^2-4ac##) must be negative. That will give you some constraints on ##\bar{BA}## which you can then use to work out the minimum perimeter triangle.$$\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

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.