<p>\(T_1\) is an isosceles triangle with circumcircle K. Let \(T_2\) be another isosceles triangle inscribed in K whose base is one of the equal side of \(T_1\) and which overlaps the interior of \(T_1\). Similarly create isosceles triangles \(T_3\) from \(T_2\), \(T_4\) from \(T_3\) and so on to the triangle \(T_n\). Then the base angle of the triangle \(T_n\) as \(n \to \infty\) is</p>
Step-by-Step Solution
Key Concept: Each successive isosceles triangle is constructed with its base as the equal side of the previous triangle, creating a recursive angle relationship where if the apex angle of Tₙ is θₙ, then the base angle is (π-θₙ)/2, and this generates a functional equation θₙ₊₁ = (π-θₙ)/2 that converges to a fixed point.
<p><strong>Step 1:</strong> Let αₙ be the apex angle and βₙ be the base angle of triangle Tₙ. For any isosceles triangle: αₙ + 2βₙ = π, so βₙ = (π-αₙ)/2.</p><p><strong>Step 2:</strong> The base of Tₙ (length determined by apex angle αₙ) becomes the equal side of Tₙ₊₁. By the inscribed angle geometry in circle K, the apex angle of Tₙ₊₁ equals the base angle of Tₙ. Therefore: αₙ₊₁ = βₙ = (π-αₙ)/2.</p><p><strong>Step 3:</strong> As n → ∞, let αₙ → L. Then L satisfies: L = (π-L)/2, which gives 2L = π - L, so 3L = π, thus L = π/3.</p><p><strong>Step 4:</strong> The limiting base angle is β = (π-L)/2 = (π-π/3)/2 = (2π/3)/2 = π/3.</p><p>∴ Answer: B (base angle → π/3 or 60°)</p>
Correct Answer: B