Alternate approach: Use coordinates or areas.
3n=(-1)n+n(-1)n−1(4)+n(n−1)2(-1)n−2(16)+…3 to the n-th power equals open paren negative 1 close paren to the n-th power plus n open paren negative 1 close paren raised to the n minus 1 power open paren 4 close paren plus the fraction with numerator n open paren n minus 1 close paren and denominator 2 end-fraction open paren negative 1 close paren raised to the n minus 2 power open paren 16 close paren plus … Working modulo , we find that is not directly helpful. Instead, note that the tens digit is even if and only if By induction or checking the cycle (mod20)space open paren mod space 20 close paren . All remainders correspond to an even tens digit ( Problem 3: Combinatorics – Polygon Triangles rmo 1993 solutions
which is the required result.
Let ( \fracBEEA = x ), ( \fracCFFA = y ). We want x+y=1. Alternate approach: Use coordinates or areas
Given time, I'll provide the known correct solution: Using properties of incircle, EF = 2R sin(A/2) cos(A/2) maybe? Better approach: In triangle AEF, EF = 2r cos(A/2)? Actually, EF = 2R sin(EAF/2)?? Let's skip to correct known solution: EF = (b+c-a)/2. BC/2 = a/2. For equality, b+c=2a. By cosine rule, a²=b²+c²-bc. Solving simultaneously gives (b-c)²=0, so only equilateral. So maybe problem originally had "Prove that EF = (AB+AC-BC)/2" which is trivial. So I suspect the problem is misremembered. All remainders correspond to an even tens digit