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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 780.
Let D, E, F are the mid-points of BC, CA, AB respectively. Then1/2*a*OD + 1/2*b*OE + 1/2*c*OF= Area of triangle ABC= 1/2*(a+b+c)*r ....... (1)Also, from OEAF, OFBD, ODCE concyclic, and by Ptolemy theorem, R*a/2 = OF*b/2 + OE*c/2R*b/2 = OF*a/2 + OD*c/2R*c/2 = OE*a/2 + OD*b/2Summing up, 1/2*(a+b+c)*R= a/2*(OE+OF) + b/2*(OD+OF) + c/2*(OD+OE) ....... (2)Now (1)+(2), 1/2*(a+b+c)*(R+r)= 1/2*(a+b+c)*(OD+OE+OF)Thus, OD+OE+OF = R+rOn the other hand, HA = 2*ODHB = 2*OEHC = 2*OFHence, HA + HB + HC = 2*(R+r)
cos A + cos B + cos C = 1 + 4 sin A/2 . sin B/2 . sin C/2=> 2R cos A + 2R cos B + 2R cos C = 2R + 2(4R sin A/2 . sin B/2 . sin C/2)=> HA + HB + HC = 2R + 2r = 2(R + r)
It´s easy to express AH in terms of A, B, b as AH = b*cos(A)/sin(B), from extended Sine Law we can see that a/sin A = b/sin B = c/sin C = 2R, therefore AH = 2R*cos A, we gain BH and CH similarily, we obtain AH+BH+CH = 2R(cos(A)+cos(B)+cos(C)). The immediate consequence of Carnot´s theorem is that cos(A) + cos(B) + cos(C) = 1 + r/R. Using that we obtain the result.