Talk:Apollonius's theorem: Difference between revisions
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==Reduction to Pythagoras== |
==Reduction to Pythagoras== |
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While the current formulation is fine, the criticism of earlier formulation by the IP was imho not correct. The original formulation was correct as well if you look at it the right way (see associated drawing in the article). Note that strictly speaking in both cases it only "essentially" reduces to Pythagoras as in both cases you still need minimal rearrangement. In the case of ABC having a right angle you need to move 2 green squares to see that it is Pythagoras and in the case of the isosceles ABC you need to divide the theorem's equation by two, so that that yields the Pythagoras for ABD. However, personally I liked the old version better though, as the whole point of seeing Apollonius' or Stewart's theorem as a generalisation of Pythagoras is that they yield you Pythagoras for the original triangle, that is ABC rather than ABD.--[[User:Kmhkmh|Kmhkmh]] ([[User talk:Kmhkmh|talk]]) 05:37, 30 May 2017 (UTC) |
While the current formulation is fine, the criticism of earlier formulation by the IP was imho not correct. The original formulation was correct as well if you look at it the right way (see associated drawing in the article). Note that strictly speaking in both cases it only "essentially" reduces to Pythagoras as in both cases you still need minimal rearrangement. In the case of ABC having a right angle you need to move 2 green squares to see that it is Pythagoras and in the case of the isosceles ABC you need to divide the theorem's equation by two, so that that yields the Pythagoras for ABD. However, personally I liked the old version better though, as the whole point of seeing Apollonius' or Stewart's theorem as a generalisation of Pythagoras is that they yield you Pythagoras for the original triangle, that is ABC rather than ABD.--[[User:Kmhkmh|Kmhkmh]] ([[User talk:Kmhkmh|talk]]) 05:37, 30 May 2017 (UTC) |
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== What is Apollonius' proof? == |
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The proof given uses trigonometry, which was not a tool available to Apollonius. So, how did he prove it? |
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Revision as of 04:50, 9 July 2020
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Stewart's theorem?
I think that Apollonius' theorem is the spacial case m=n=1. or else' what is the diference between Apollonius' theorem & Stewart's theorem?! It'sthe same theorem!!!!
- Most of the sources I looked at use Apollonius as the name for the median case and Stewart for the general case. And I've modified the article to reflect that. MathWorld indicates they are actually synonyms but I couldn't find an independent source for this.--RDBury (talk) 20:16, 1 April 2010 (UTC)
1. I note that planetmath is given as a reference, but the theorem as given in planetmath is wrong: b² + c² = 2m² + [a² /2] - where in their notation, m is the length of the median. Using the notation in the (confusingly labelled*) figure on this Wikipedia page, where d is the length of the median and m is each of the two halves of a, planetmath would read:
b² + c² = 2d² + (a²/2) or using the actual formulation on the Wikipedia page:
b² + c² = 2d² + (2m² /2)
This is clearly not the same as the correct formula on the Wikipedia page: b² + c² = 2d² + 2m², they have got muddled and divided a²by 2 as well as multiplying. Planetmath is not a useful reference.
- it is not immediately obvious whether Wikipedia's letter a is intended to be the side a opposite the vertex A or is meant to be the point where the median meets BC.
2. However the derivation in the first half of the Wikipedia page from the cosine formula omits a² in the third equation and, also in that equation, changes the sign of the last term from - to +: so the derivation does not work.
Can I suggest that this page and the references be revised ? Ardj (talk) 17:50, 1 September 2013 (UTC)
Proof needs to be edited.
The line before "_Add these equations to obtain_" misses something befor the 'equals sign'. — Preceding unsigned comment added by 27.255.241.149 (talk) 15:15, 16 July 2015 (UTC)
Reduction to Pythagoras
While the current formulation is fine, the criticism of earlier formulation by the IP was imho not correct. The original formulation was correct as well if you look at it the right way (see associated drawing in the article). Note that strictly speaking in both cases it only "essentially" reduces to Pythagoras as in both cases you still need minimal rearrangement. In the case of ABC having a right angle you need to move 2 green squares to see that it is Pythagoras and in the case of the isosceles ABC you need to divide the theorem's equation by two, so that that yields the Pythagoras for ABD. However, personally I liked the old version better though, as the whole point of seeing Apollonius' or Stewart's theorem as a generalisation of Pythagoras is that they yield you Pythagoras for the original triangle, that is ABC rather than ABD.--Kmhkmh (talk) 05:37, 30 May 2017 (UTC)
What is Apollonius' proof?
The proof given uses trigonometry, which was not a tool available to Apollonius. So, how did he prove it?