Analytic Geometry – Problem 12: Find the maximum value of

In triangle $ABC$, $a^2+b^2+c^2=2\sqrt3\,bc\sin A$, $a=6$, and point $D$ satisfies $DC=2DB$. Find the maximum value of $DA$.

Given $$a^2+b^2+c^2=2\sqrt3\,bc\sin A,\qquad a=6.$$ From cosine law, $$a^2=b^2+c^2-2bc\cos A.$$ Substitute into the given relation: $$2(b^2+c^2)-2bc\cos A=2\sqrt3\,bc\sin A,$$ so $$b^2+c^2=bc(\cos A+\sqrt3\sin A)=2bc\sin\!\left(A+\frac{\pi}{6}\right).$$ Since $b^2+c^2\ge2bc$ and $\sin(A+\pi/6)\le1$, we must have equality in both: $$b^2+c^2=2bc,\quad \sin\!\left(A+\frac{\pi}{6}\right)=1.$$ Thus $b=c$, $A=\pi/3$, and triangle $ABC$ is equilateral with side $6$.

Choose coordinates $$B(-3,0),\quad C(3,0),\quad A(0,3\sqrt3).$$ Let $D(x,y)$. Condition $DC=2DB$ gives $$(x-3)^2+y^2=4\bigl((x+3)^2+y^2\bigr),$$ which simplifies to $$(x+5)^2+y^2=16.$$ So $D$ lies on the circle centered at $(-5,0)$ with radius $4$.

Hence $$DA_{\max}=|A(-5,0)|+4=\sqrt{(0+5)^2+(3\sqrt3)^2}+4=\sqrt{25+27}+4=4+2\sqrt{13}.$$

$$DA_{\max}=4+2\sqrt{13}.$$