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2024/25 AHSMC Part 1 Problem 8

Problem 8 of 16

AA, BB, CC, FF are points on a straight line (in this order). ABAB is the diameter of a circle, and the line CECE is tangent to the circle at point DD, where DD is between CC and EE. If ∠DCF=2(∠ADE−∠BAD)\angle DCF = 2(\angle ADE - \angle BAD), find the value of ∠ADE\angle ADE.

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Solution

Let α=∠ACD\alpha = \angle ACD, β=∠BAD\beta = \angle BAD and θ=∠ADE\theta = \angle ADE. Then ∠BDC=90∘−θ\angle BDC = 90^\circ - \theta. In △BCD\triangle BCD, ∠CBD=180∘−α−(90∘−θ)=90∘−α+θ, \angle CBD = 180^\circ - \alpha - (90^\circ - \theta) = 90^\circ - \alpha + \theta, so ∠ABD=90∘+α−θ\angle ABD = 90^\circ + \alpha - \theta. Since ∠BAD+∠ABD=90∘\angle BAD + \angle ABD = 90^\circ, 90∘=90∘+α−θ+β  ⟹  θ=α+β. 90^\circ = 90^\circ + \alpha - \theta + \beta \implies \theta = \alpha + \beta. The given condition is 180∘−α=2(θ−β)=2α180^\circ - \alpha = 2(\theta - \beta) = 2\alpha, so α=60∘\alpha = 60^\circ. If OO is the centre then OD⊥DEOD \perp DE, whence ∠COD=30∘\angle COD = 30^\circ and ∠DOA=150∘\angle DOA = 150^\circ. As △ODA\triangle ODA is isosceles, β=15∘\beta = 15^\circ and θ=60∘+15∘=75∘\theta = 60^\circ + 15^\circ = 75^\circ. The answer is (E).

University of Alberta