Inequality Involving Triangle Area and Sides
Date: 05/11/2003 at 21:14:26 From: Bobby Subject: Area of triangles Prove that for any triangle with sides a, b, and c and area A, a^2+b^2+c^2 greater or equal to (=>) 4sqrt(3)A I have found for an equilateral triangle that the inequality is equal to 4sqrt(3)A. I want to show that for an isosceles triangle and right triangle the inequality is greater than 4sqrt(3)A. In the equilateral case a=b=c all fit nicely. Should I use the same system as I found in the equilateral case, or should I be using another method?
Date: 05/12/2003 at 09:54:34 From: Doctor Floor Subject: Re: Area of triangles Hi, Bobby, Thanks for your question. To avoid misunderstanding between angle A and area, I write Area. From the law of cosines we know: -a^2 + b^2 + c^2 = 2bc cosA = 2bc sinA * cosA/sinA = 4Area * cotA We may find similar formulas with cotB and cotC. Adding these gives a^2 + b^2 + c^2 = 4Area * (cotA + cotB + cotC) Now let us write S = (a^2+b^2+c^2)/(4Area) then S = cotA + cotB + cotC (in fact, S is the cotangent of Brocard's angle, but that is unimportant for the rest of this message). Now from the following message from the Dr. Math archives: Deriving a Trig Identity http://mathforum.org/library/drmath/view/54068.html we see that in a triangle the identity tanA + tanB + tanC = tanA tanB tanC holds, or by division through by tanA tanB tanC cotB cotC + cotA cotC + cotA cotB = 1. By squaring both sides of S = cotA + cotB + cotC we find S^2 = (cotA)^2 + (cotB)^2 + (cotC)^2 + 2. Now we know that (cotA - cotB)^2 + (cotB - cotC)^2 + (cotC - cotA)^2 >= 0 and thus 2((cotA)^2 + (cotB)^2 + (cotC)^2) - 2(cotB cotC + cotA cotC + cotA cotB) >=0 or 2(S^2 - 2) - 2 >= 0 2S^2 - 6 >= 0 S^2 - 3 >= 0 and we find that S>= sqrt(3) (as it must be positive), which is exactly what you were looking for. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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