A Proof for Strong version of Ptolemy - You can use law of Cosines in ABC and ADC and use the fact that opposite angles in a cyclic quadrilateral sum to 180 Therefore you can eliminate cos terms by manipulations and then factor out the numerator to get the same expression as shown in video
Yes, there are several number theoretical proofs to this problem. They are all rather technical, however. You can check some of them out here artofproblemsolving.com/community/c6h17474p119217
ab+cd>ac+bd ab-ac>bd-cd a(b-c)>d(b-c) since b-c>0 this is equivalent to a>d which obviously holds. Similarly for the other one we get ac+bd>ad+bc a(c-d)>b(c-d) a>b. You may want to read about something called "Rearrangement inequality" as it describes the general theory for these kinds of inequalities.
You can use law of Cosines in ABC and ADC and use the fact that opposite angles in a cyclic quadrilateral sum to 180 Therefore you can eliminate cos terms by manipulations and then factor out the numerator to get the same expression
![IMO, a very Cool Inequality [ International Math Olympiad Problem ]](http://i.ytimg.com/vi/hrFeSQZrjgI/mqdefault.jpg)
A Proof for Strong version of Ptolemy - You can use law of Cosines in ABC and ADC and use the fact that opposite angles in a cyclic quadrilateral sum to 180
Therefore you can eliminate cos terms by manipulations and then factor out the numerator to get the same expression as shown in video
To your knowledge were there non geometric solutions to the problem?
Yes, there are several number theoretical proofs to this problem. They are all rather technical, however. You can check some of them out here artofproblemsolving.com/community/c6h17474p119217
7:37 It's not clear to me how we know that ab+cd>ac+bd>ad+bc?
ab+cd>ac+bd ab-ac>bd-cd a(b-c)>d(b-c) since b-c>0 this is equivalent to a>d which obviously holds. Similarly for the other one we get ac+bd>ad+bc a(c-d)>b(c-d) a>b.
You may want to read about something called "Rearrangement inequality" as it describes the general theory for these kinds of inequalities.
> which obviously holds
I’m lost.
Edit: never mind I substituted (c-d) as s and it made more sense for a sec
Not really related to the problem ,but can you prove that ab+bc+cd+da>=2ac+2bd?
Edit:if a>b>c>d
But since alpha and beta are 60. xyw and zyw are equilateral triangles. So a=c and b=d which contradicts the inequality that a>b>c>d.
Thanks for the nice explanation. One question though, how did you claim WY is an integer just because b and d are integers?
Because he was refering to WY² since WY is the squared root of an integer, then WY² it's simply an integer.
Euclidean geometry (Evan chen) solved this by geometry
i think i missed out somewhere... how is the strong version of the formula derived?
You can use law of Cosines in ABC and ADC and use the fact that opposite angles in a cyclic quadrilateral sum to 180
Therefore you can eliminate cos terms by manipulations and then factor out the numerator to get the same expression