Trigonometry & Inverse Trigonometry
Triangle Properties
Grade 11
Question:
<p>If H is the orthocentre of triangle ABC, R = circumradius and P = AH + BH + CH, then</p>
<p>(a) P = 2(R + r)</p>
<p>(b) max. of P is 3R</p>
<p>(c) min. of P is 3R</p>
<p>(d) P = 2(R - r)</p>
Step-by-Step Solution
Key Concept: Express distances from vertices to orthocentre in terms of circumradius and angles, then optimize the sum.
<p>Using the property that for any triangle, \(AH = 2R\cos A\), \(BH = 2R\cos B\), \(CH = 2R\cos C\):</p><p>\(P = 2R(\cos A + \cos B + \cos C)\)</p><p>For any triangle, \(\cos A + \cos B + \cos C \leq \frac{3}{2}\), with equality when \(A = B = C = 60°\)</p><p>Therefore \(\max(P) = 2R \cdot \frac{3}{2} = 3R\)</p>
Correct Answer: B