All Big Issues
The Instigator
Pro (for)
Losing
6 Points
The Contender
Con (against)
Winning
21 Points

# ODT2: All triangles are isosceles

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after 8 votes the winner is...
 Voting Style: Open Point System: 7 Point Started: 11/10/2012 Category: Science Updated: 5 years ago Status: Post Voting Period Viewed: 5,361 times Debate No: 27092
Debate Rounds (5)

15 comments have been posted on this debate. Showing 1 through 10 records.
Posted by Cometflash 5 years ago
I guess everyone that voted, was undecided, if all triangles are isosceles or not, since everyone vote as tie.

You guys probably kept the confusing to the voters, because after, they were still undecided, since again they all voted tie.
Posted by royalpaladin 5 years ago
My challenge was to spot the flaw in the proof.
Posted by Luggs 5 years ago
You do realize that all that needs to be done to prove Pro wrong is to measure the sides. By definition, an isosceles triangle has at least 2 sides that are equivalent, and the angles opposite those sides have equivalent are equivalent as well. Therefore, to prove that not all triangles are isosceles, you only need to measure sides and angles (better to measure them all to make sure).
Posted by staraa 5 years ago
though the topic is quite interesting but you can't prove it true for scalene triangle. can you?????
if yes then you are really a genius
Posted by baggins 5 years ago
@ Costas

Redefining the metric would have been a nice trick. However this proof is constrained to high school geometry (i.e R2 with the usual metric).
Posted by costas 5 years ago
There is a specific step of the proof that is false. So I think Con should focus on spotting this flaw instead of trying to make some general argument against the proof.

By the way, since there is no reference to the structure of the space where the triangles lie, one could assume they are on R^2 with the trivial metric d((x,x'),(y,y'))=1, for all real x,x',y,y'. Hence all triangles are isosceles and furthermore they are equilateral.

I guess Pro assumes they are in R^2 with the usual metric d((x,x'),(y,y'))=sqrt((x-y)^2+(x'-y')^2).
Posted by baggins 5 years ago
The proof is intentionally verbose. The idea was to create a bit of panic...
Posted by Zaradi 5 years ago
*looks at debate*