IMO/KKK/Geometric Inequality/1 Geometric Inequalities

Notation and Basic Facts a, b, and c are the sides of ∆ABC opposite to A, B, and C respectively. [ABC] = area of ∆ABC s = semi-perimeter =) c b a (2 1 + + r = inradius R = circumradius Sine Rule: R 2 C sin c B sin b A sin a = = = Cosine Rule: a 2 = b 2 + c 2 − 2bc cos A [ABC] = B sin ac 2 1 A sin bc 2 1 C sin ab 2 1 = = = R 4 abc =) c s)(b s)(a s (s − − − (Heron's Formula) = 2 cr 2 br 2 ar + + = sr Example 1 Isoperimetric Theorem for Triangle Among all triangles with a fixed perimeter, the equilateral triangle has the largest area. Proof: Using the Heron's Formula and the AM-GM inequality [ABC] = 3 3 s 3 s s 3) c s () b s () a s (s) c s)(b s)(a s (s 2 3 3 =       =       − + − + − ≤ − − − with equality holds if and only if s − a = s − b = s− c, i.e a = b = c. Example 2 [IMO 1961] Let a, b, c be the sides of a triangle, and T its area. Prove: a 2 + b 2 + c 2 ≥ 4 3 T. In what case does equality hold? 1st solution: Denote the perimeter of the triangle by p, i.e p = a + b + c, by the isoperimetric theorem for triangle, we have T ≤ 4 3 3 p 2       with equality holds if and only if a = b = c-(1) The Cauchy-Schwarz inequality gives p 2 = (a + b + c) 2 ≤ 3(a 2 + b 2 + c 2) with equality holds if and only if a = b = c.-(2) It follows from (1) and (2) that T ≤ 4 3 3 c b a 2 2 2 + + which is equivalent to a 2 + b 2 + c 2 ≥ 4 3 T with equality holds if and only if a = b = c. 2nd solution: An equilateral triangle with side c has altitude 2 3 …