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let AABC be a triangle S^2. Let the dual point C'ES^2 of C be defined by the following three conditions: (i) d(C¹, A)=π/2 (ii) d(C', B)=π/2 (iii) d(C¹, C)<π/2 A' and B' are defined analogously. Thus we get a dual triangle AA' B' C'. More precisely, AABC is a non- degenerate triangle, and it follows (you may assume) that AA' B' C' is too. (a) Are there triangles AABC in S^2 identical to their own dual AA'B'C'?

let AABC be a triangle S^2. Let the dual point C'ES^2 of C be defined by the following three conditions: (i) d(C¹, A)=π/2 (ii) d(C', B)=π/2 (iii) d(C¹, C)<π/2 A' and B' are defined analogously. Thus we get a dual triangle AA' B' C'. More precisely, AABC is a non- degenerate triangle, and it follows (you may assume) that AA' B' C' is too. (a) Are there triangles AABC in S^2 identical to their own dual AA'B'C'?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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et △ABC be a triangle in S^2 (the two-dimensional sphere). Let the dual point C'∈S^2 of C be defined by the following three conditions:
(i) d(C' , A)=π/2
(ii) d(C' , B)=π/2
(iii) d(C' , C)≤π/2
A' and B' are defined analogously. Thus we get a dual triangle △A' B' C'. More precisely, △ABC is a non-degenerate triangle, and
it follows (you may assume) that △A' B' C' is too.
(question) Are there triangles △ABC in S^2 identical to their own dual △A'B'C'?

 

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let AABC be a triangle S^2. Let the dual point C'ES^2 of C be defined by the following three conditions: (i) d(C¹, A)=π/2 (ii) d(C¹, B)=π/2 (iii) d(C¹, C)<π/2 A' and B' are defined analogously. Thus we get a dual triangle AA' B' C'. More precisely, AABC is a non- degenerate triangle, and it follows (you may assume) that AA' B' C' is too. (a) Are there triangles AABC in S^2 identical to their own dual AA'B'C'?
Transcribed Image Text:let AABC be a triangle S^2. Let the dual point C'ES^2 of C be defined by the following three conditions: (i) d(C¹, A)=π/2 (ii) d(C¹, B)=π/2 (iii) d(C¹, C)<π/2 A' and B' are defined analogously. Thus we get a dual triangle AA' B' C'. More precisely, AABC is a non- degenerate triangle, and it follows (you may assume) that AA' B' C' is too. (a) Are there triangles AABC in S^2 identical to their own dual AA'B'C'?
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