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| Sixty is Special (Posted on 2017-08-18) |
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I is the incenter of triangle ABC with
B' the intersection of the ray BI and
side AC and C' the intersection of ray
CI and side AB.
Prove that AB'IC' is a cyclic quadrilateral if and only if angle BAC is 60 degrees.
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Submitted by Bractals
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Solution:
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I first tried to prove that AB'IC' was cyclic if and only if the
point I lies on the circumcircle of ΔAB'C', but that started to
generate too much algebra.
The following is more geometrical and the difficulty level
drops from 4 to 2.
Let α, β, and γ be the measures of ∠BAC, ∠CBA, and ∠ACB
respectively. ∠AB'I and ∠AC'I are exterior angles of ΔB'CB
and ΔC'BC respectively. Thus,
∠AB'I = ∠B'CB + ∠CBB' = γ + β/2 and
∠AC'I = ∠C'BC + ∠BCC' = β + γ/2 and
Therefore,
α = 60° <==> β + γ = 120°
<==> 3*(β + γ)/2 = 180°
<==> (γ + β/2) + (β + γ/2) = 180°
<==> ∠AB'I + ∠AC'I = 180°
<==> AB'IC' is cyclic.
QED
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