All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Sixty is Special (Posted on 2017-08-18) Difficulty: 4 of 5
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.

  Submitted by Bractals    
No Rating
Solution: (Hide)

  
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
  

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolution & a further resultJer2017-08-27 17:58:50
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information