The 80-80-20 Triangle Problem, Solution #8
Let ABC be an isosceles triangle
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Copyright © 1996-2018 Alexander Bogomolny
This is Solution 7 from [Knop] and is due to Alexander Kornienko, a high school student at the time.
Reflect B in AC and C in AB to obtain B' and, respectively, C'.
Now, ∠AB'D = ∠ABD = 30°. Further,
Since ∠AC'E = ∠ACE = ∠CAE = ∠C'AE, we see that E is equidistant from A, C, and C'. E then lies on the perpendicular bisector of AC', i.e. on B'D. But then
∠CED = 180° - 40° - 40° - 70° = 30°.
(This solution is neatly embedded into a configuration of an 18-gon.)
Reference
- C. Knop, Nine Solutions to One Problem, Kvant, 1993, no 6.
The 80-80-20 Triangle Problem
- Solution #1
- Solution #2
- Solution #3
- Solution #4
- Solution #5
- Solution #6
- Solution #7
- Solution #8
- Solution #9
- Solution #10
- Solution #11
- Solution #12
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Copyright © 1996-2018 Alexander Bogomolny
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