<p>If the altitudes of triangle ABC are in harmonic progression, then the side length $b$ (which is CA) can be</p>
Step-by-Step Solution
Key Concept: If altitudes are in harmonic progression, then their reciprocals (which are proportional to the sides) are in arithmetic progression. Use the relationship between altitudes and area to find constraints on the sides.
<p><strong>Step 1:</strong> Express altitudes in terms of sides and area. For triangle ABC with area Δ: h_a = 2Δ/a, h_b = 2Δ/b, h_c = 2Δ/c, where a, b, c are sides opposite to vertices A, B, C respectively.</p><p><strong>Step 2:</strong> Given that altitudes h_a, h_b, h_c are in H.P., their reciprocals are in A.P.: 1/h_a, 1/h_b, 1/h_c are in A.P.</p><p><strong>Step 3:</strong> Since 1/h_a = a/(2Δ), 1/h_b = b/(2Δ), 1/h_c = c/(2Δ), this means: a/(2Δ), b/(2Δ), c/(2Δ) are in A.P.</p><p><strong>Step 4:</strong> Simplifying: a, b, c are in A.P. Therefore: 2b = a + c.</p><p><strong>Step 5:</strong> By the triangle inequality, we need a + c > b. Since 2b = a + c, we have 2b > b, giving b > 0 (always true). Also, b + c > a and b + a > c must hold.</p><p><strong>Step 6:</strong> From 2b = a + c: if a = 2 and c = 4, then 2b = 6, so b = 3. Check triangle inequality: 2 + 4 > 3 ✓, 2 + 3 > 4 ✓, 3 + 4 > 2 ✓. All conditions satisfied.</p><p><strong>Step 7:</strong> Testing option (Q) b = 3: We can have sides a = 2, b = 3, c = 4 in A.P., which means altitudes are in H.P.</p><p><strong>∴ Answer: Q</strong></p>
Correct Answer: Q