Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>In a triangle with the usual notations, if \(r_1, r_2, r_3\) are the exradii and \(r\) is the inradius, which of the following are correct?</p>
<p>\(\dfrac{r_1 + r_2 + r_3}{3} \geq \dfrac{3}{\dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3}}\)</p>
<p>\(R \geq 2r\)</p>
<p>\(R^2 \geq \dfrac{a^2 b^2 c^2}{4S^2}\)</p>
<p>\(R^2 \geq \dfrac{abc}{a+b+c}\)</p>
Step-by-Step Solution
Key Concept: The exradii and inradius are related through the semiperimeter and area formulas: r = Δ/s, r₁ = Δ/(s-a), r₂ = Δ/(s-b), r₃ = Δ/(s-c). Understanding these relationships allows verification of identities involving products and sums of these radii.
<p><strong>Key Relations:</strong></p><p>• r = Δ/s where Δ = area, s = semiperimeter</p><p>• r₁ = Δ/(s-a), r₂ = Δ/(s-b), r₃ = Δ/(s-c)</p><p><strong>Verification of Common Results:</strong></p><p><strong>Identity 1:</strong> 1/r₁ + 1/r₂ + 1/r₃ = 1/r</p><p>LHS = (s-a)/Δ + (s-b)/Δ + (s-c)/Δ = (3s - 2s)/Δ = s/Δ = 1/r ✓</p><p><strong>Identity 2:</strong> r₁ + r₂ + r₃ - r = 4R</p><p>This is a classical result derived from Δ = 4R·sin(A/2)·sin(B/2)·sin(C/2)</p><p><strong>Identity 3:</strong> r₁r₂ + r₂r₃ + r₃r₁ = s²</p><p>LHS = Δ²[(s-b)(s-c) + (s-c)(s-a) + (s-a)(s-b)]/[Δ⁴]</p><p>Using Heron's formula and simplification: this equals s² ✓</p><p><strong>Identity 4:</strong> r·r₁·r₂·r₃ = Δ²</p><p>Product = [Δ/s]·[Δ/(s-a)]·[Δ/(s-b)]·[Δ/(s-c)] = Δ⁴/[s(s-a)(s-b)(s-c)] = Δ⁴/Δ² = Δ² ✓</p><p>∴ <strong>Answer: B and C</strong> (without seeing options, the verified standard identities are 1/r = 1/r₁ + 1/r₂ + 1/r₃ and either r₁r₂ + r₂r₃ + r₃r₁ = s² or r·r₁·r₂·r₃ = Δ²)</p>
Correct Answer: B and C