Question

Let ABC be an acute-angled triangle with area R. Then \( \sqrt{a^2b^2-4R^2} + \sqrt{b^2c^2-4R^2} + \sqrt{c^2a^2-4R^2} = \)

• A. \( a+b+c \)
• B. \( a^2+b^2+c^2 \)
• C. \( \frac{a^2+b^2+c^2}{2} \)
• D. \( 2(a^2+b^2+c^2) \)

Hint:

Area of triangle \(\Delta = \frac{1}{2}ab\sin C\). So \(2\Delta = ab\sin C\).
\(4\Delta^2 = a^2b^2\sin^2 C\).
The term \(\sqrt{a^2b^2-4\Delta^2} = \sqrt{a^2b^2(1-\sin^2 C)} = \sqrt{a^2b^2\cos^2 C} = |ab\cos C|\).
For acute triangle, \(\cos C>0\), so it's \(ab\cos C\).
Law of Cosines: \(c^2 = a^2+b^2-2ab\cos C \Rightarrow ab\cos C = (a^2+b^2-c^2)/2\).

Answer 1

The Correct Option is C

Let the area of \(\triangle ABC\) be denoted by \(\Delta\). The question uses R for area, so \(\Delta = R\). We know that the area of a triangle can also be expressed as \(\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B\). So, \(2\Delta = ab\sin C\). Squaring this: \(4\Delta^2 = a^2b^2\sin^2 C\). The term under the first square root is \(a^2b^2-4R^2 = a^2b^2-4\Delta^2\). Substitute \(4\Delta^2 = a^2b^2\sin^2 C\): \(a^2b^2 - 4\Delta^2 = a^2b^2 - a^2b^2\sin^2 C = a^2b^2(1-\sin^2 C) = a^2b^2\cos^2 C\). So, \(\sqrt{a^2b^2-4\Delta^2} = \sqrt{a^2b^2\cos^2 C} = |ab\cos C|\). Since ABC is an acute-angled triangle, all angles A, B, C are less than \(90^\circ\), so \(\cos A, \cos B, \cos C\) are all positive. Thus, \(\sqrt{a^2b^2-4\Delta^2} = ab\cos C\). Similarly, \(\sqrt{b^2c^2-4\Delta^2} = bc\cos A\). \(\sqrt{c^2a^2-4\Delta^2} = ca\cos B\). The expression becomes \(ab\cos C + bc\cos A + ca\cos B\). Using the Law of Cosines: \(\cos C = \frac{a^2+b^2-c^2}{2ab} \Rightarrow ab\cos C = \frac{a^2+b^2-c^2}{2}\). \(\cos A = \frac{b^2+c^2-a^2}{2bc} \Rightarrow bc\cos A = \frac{b^2+c^2-a^2}{2}\). \(\cos B = \frac{c^2+a^2-b^2}{2ca} \Rightarrow ca\cos B = \frac{c^2+a^2-b^2}{2}\). Summing these three terms: Expression = \( \frac{a^2+b^2-c^2}{2} + \frac{b^2+c^2-a^2}{2} + \frac{c^2+a^2-b^2}{2} \) \( = \frac{1}{2} [ (a^2+b^2-c^2) + (b^2+c^2-a^2) + (c^2+a^2-b^2) ] \) \( = \frac{1}{2} [ a^2+b^2-c^2 + b^2+c^2-a^2 + c^2+a^2-b^2 ] \) In the numerator, \(a^2-a^2+a^2 = a^2\), \(b^2+b^2-b^2 = b^2\), \(-c^2+c^2+c^2 = c^2\). So, Numerator = \(a^2+b^2+c^2\). Expression = \( \frac{a^2+b^2+c^2}{2} \). This matches option (c). \[ \boxed{\frac{a^2+b^2+c^2}{2}} \]