[Geometry]
Problem: In triangle $A B C, A B=4, B C=6$, and $A C=8$. Squares $A B Q R$ and $B C S T$ are drawn external to and lie in the same plane as $\triangle A B C$. Compute $Q T$.

Solution: Set $\mathrm{m} \angle A B C=x$ and $\mathrm{m} \angle T B Q=y$. Then $x+y=180^{\circ}$ and so $\cos x+\cos y=0$. Applying the Law of Cosines to triangles $A B C$ and $T B Q$ gives $A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos x$ and $Q T^{2}=B T^{2}+B Q^{2}-2 B T \cdot B Q \cos y$, which, after substituting values, become $8^{2}=$ $4^{2}+6^{2}-48 \cos x$ and $Q T^{2}=4^{2}+6^{2}-48 \cos y$.

Adding the last two equations yields $Q T^{2}+8^{2}=2\left(4^{2}+6^{2}\right)$ or $Q T=\mathbf{2} \sqrt{\mathbf{1 0}}$.

Final answer: $2 \sqrt{10}$