| Posted on Tuesday, 14 October, 2003 - 02:53 am: |
triangle ABC has circumradius 1
any point M on the circumcircle, MA*MB*MC< =2
show it is equilateral
| Posted on Tuesday, 14 October, 2003 - 02:50 pm: |
this can probably be done by using the expression for a general point on a circumcircle, and multiplying the generalised expressions for MA,MB,MC, perhaps using vectors.
To prove it, you should end with a resultant equation that shows for all other triangles the product is greater than 2.
You may also be able to use RAA.
| Posted on Tuesday, 14 October, 2003 - 04:17 pm: |
> To prove it, you should end with a resultant equation that shows for all other triangles the product is greater than 2.
I don't think it can be that simple, since if M = A, then the expression is zero, and thus < 2. The question is whether or not there is some point on the circumcircle for which MA*MB*MC > 2. You can attack this either by guessing a suitable point and showing that if A is not equilateral it has the required property, or by finding explicitly the point at which MA*MB*MC is maximised.
David
| Posted on Tuesday, 14 October, 2003 - 05:04 pm: |
You could try using the fact that if M lies between A and C say on the circumcentre that MB= MA+MC ...
| Posted on Tuesday, 14 October, 2003 - 05:09 pm: |
I think if you assume a > b ³ c and take M to be the midpoint of the arc between B and C on the opposite side from A, then if D ABC is not equilateral, then it all works. In fact, if I am not mistaken in my calculations, for this M we have MA×MB×MC=8 sin2 A/2 cos(B-C)/2 = 4 sin A/2 (cos B+cos C). Since A > B ³ C, we have A > p/3, so sin A/2 > 1/2; thus it suffices to show that cos B + cos C > 1. This looks like it ought to be easy, but I haven't found a sensible proof. If we use the formula for cosine in terms of the side lengths we must show b(a2+c2-b2)+c(a2+b2-c2) > 2a b c Expanding everything out we must show a2 b + b c2 + a2 c + b2 c > 2 a b c + b3 + c3 We know that a2 b > b3 and b c2 > c3 so it suffices to show that a2 c + b2 c > 2a b c which is just c(a-b)2 > 0 and is consequently true. So we are done. (Phew!) I'll leave you the fun of checking that MA * MB * MC is never > 2 for an equilateral triangle (use Ptolemy's theorem). David
| Posted on Tuesday, 14 October, 2003 - 05:11 pm: |
Rachel, what you say is only true for an equilateral triangle (for a general triangle the corresponding statement is that a MA + c MC = b MB, which is a restatement of Ptolemy's theorem).
David
| Posted on Tuesday, 14 October, 2003 - 11:43 pm: |
A brute force solution, although solves the problem, is uninteresting...
but then again, i do not see any elegant way of attacking this problem
| Posted on Tuesday, 14 October, 2003 - 11:51 pm: |
Are you criticising my solution?
No, I agree, it is not particularly inspiring. If I get round to it I might have another think and see if I can work out the real reason why it is true - finding the point Mmax which maximises the product in the statement of the problem could be illuminating.
David
| Posted on Wednesday, 15 October, 2003 - 04:07 am: |
at least you got a solution ;)