If the square of the diameter of a circle circumscribing a $\triangle ABC$ is equal to half the sum of the squares of its sides then $\sum\sin^2 A$ is
Step-by-Step Solution
Key Concept: Apply the sine rule to relate sides to sines of angles and use the given constraint on the circumradius.
For a triangle inscribed in a circle with circumradius $R$, we use the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Squaring both sides gives $\sin^2 A + \sin^2 B + \sin^2 C = \frac{a^2 + b^2 + c^2}{4R^2}$. From the given condition $(2R)^2 = \frac{a^2 + b^2 + c^2}{2}$, we obtain $a^2 + b^2 + c^2 = 8R^2$, which yields $\sin^2 A + \sin^2 B + \sin^2 C = 2$.
Correct Answer: 2