<p>In an equilateral △ABC (where symbols used have usual meanings), then <i>r</i>, <i>R</i> and <i>r</i><sub>1</sub> form:</p>
Step-by-Step Solution
Key Concept: For an equilateral triangle, we need to find expressions for the inradius (r), circumradius (R), and exradius (r₁), then determine their progression relationship by checking if their reciprocals form an A.P.
<p><strong>Step 1: Find r for an equilateral triangle with side a</strong></p><p>For any triangle: r = (s-a)tan(A/2), where s is semi-perimeter.</p><p>For equilateral triangle: A = 60°, s = 3a/2</p><p>r = (3a/2 - a)tan(30°) = (a/2) · (1/√3) = <strong>a/(2√3)</strong></p><p>Alternatively, r = Δ/s where Δ = (√3/4)a² and s = 3a/2</p><p>r = (√3a²/4)/(3a/2) = <strong>a√3/6</strong></p><p><strong>Step 2: Find R for an equilateral triangle</strong></p><p>Using R = a/(2sin A): R = a/(2sin 60°) = a/(2 · √3/2) = <strong>a/√3</strong></p><p>Or R = <strong>a√3/3</strong></p><p><strong>Step 3: Find r₁ for an equilateral triangle</strong></p><p>r₁ = s·tan(A/2) where s is semi-perimeter</p><p>r₁ = (3a/2)·tan(30°) = (3a/2)·(1/√3) = <strong>3a/(2√3) = a√3/2</strong></p><p><strong>Step 4: Express r, R, r₁ with common denominator</strong></p><p>r = a√3/6, R = a√3/3, r₁ = a√3/2</p><p>Factoring out a√3: r = a√3/6, R = a√3/3, r₁ = a√3/2</p><p><strong>Step 5: Check if they form an H.P. by examining reciprocals</strong></p><p>1/r = 6/(a√3), 1/R = 3/(a√3), 1/r₁ = 2/(a√3)</p><p>Dividing by common factor: 1/r : 1/R : 1/r₁ = 6 : 3 : 2</p><p><strong>Step 6: Verify if 6, 3, 2 form an A.P.</strong></p><p>Check: 3 - 6 = -3 and 2 - 3 = -1 ✗ (Not A.P. directly)</p><p>Actually: 1/6 : 1/3 : 1/2 has differences: 1/3 - 1/6 = 1/6 and 1/2 - 1/3 = 1/6 ✓</p><p>Since 1/r, 1/R, 1/r₁ form an A.P., the values r, R, r₁ form an <strong>H.P.</strong></p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C