Median Maximum (Posted on 2008-02-03)

	Submitted by Brian Smith
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Geometry/Trigonometry Solution Let the verticies of the triangle be A, B, and C. Let D be the midpoint of BC, let E be the midpoint of AC, and let F be the midpoint of AB. The medians of the triangle are then AD, BE, and CF. Let AD=15 and BE=12. The three medians intersect at a point which divides the medians into a 2 to 1 ratio. Let M be that point. Then AM=10, BM=8, DM=5, EM=4, CM=(2/3)*CF, FM=(1/3)*CF. The area of ABC is equal to 4/3 the area of trapezoid ABDE. The area of ABDE can be calculated by adding the areas of triangles ABM, BDM, DEM, and EAM. Letting T be the measure of angle AMB, the area can be calculated using the sine rule for triangle area: area ABM = (1/2)*AM*MB*sin(AMB) = (1/2)*10*8*sin(T) = 40*sin(T) area BDM = (1/2)*BM*MD*sin(BMD) = (1/2)*8*5*sin(180-T) = 20*sin(T) area DEM = (1/2)*DM*ME*sin(DME) = (1/2)*5*4*sin(T) = 10*sin(T) area EAM = (1/2)*EM*MA*sin(EMA) = (1/2)*4*10*sin(180-T) = 20*sin(T) area ABDE = 40*sin(T) + 20*sin(T) + 10*sin(T) + 20*sin(T) = 90*sin(T) area ABC = (4/3)*90*sin(T) = 120*sin(T) The area is a maximum when sin(T) is a maximum, which occurs when T=90 and sin(T)=1. Then the maximum area of triangle ABC is 120. Since angle AMB = 90, triangle AMB is a right triangle with a hypotenuse AB = sqrt(10^2+8^2) = 2*sqrt(41). Point F is the midpoint of AB and since the midpoint of the hypotenuse is the circumcenter of a right triangle, the length of FM = sqrt(41). Since FM = (1/3)*CF, this makes the length of median CF equal to 3*sqrt(41). ----- Calculus Solution The area of a triangle can be expressed in terms of the medians, similar to the formula for the sides. Let the median lengths be Ma, Mb, and Mc. Let Sm=(Ma+Mb+Mc)/2. Then the area of the triangle is: A = (4/3)*sqrt[Sm*(Sm-Ma)*(Sm-Mb)*(Sm-Mc)] Substituting the expression for Sm, and Ma=12 and Mb=15, the equation can be simplified to: A = (1/3)sqrt[(729-Mc^2)*(Mc^2-9)] Rearranging for easy differentiation: 9*A^2 = (729-Mc^2)*(Mc^2-9) (18*A)*dA/dMc = (-2Mc)*(Mc^2-9) + (729-Mc)*(2*Mc) 0 = 2*Mc*(729 - MC - MC + 9) Mc=0, +/-sqrt[369] The only positive root is Mc = sqrt[369] = 3*sqrt[41].

	Subject	Author	Date
	Using Trigonometry	Praneeth	2008-02-04 00:36:40
	another approach	Dennis	2008-02-03 13:53:18
	Solution	Bractals	2008-02-03 12:13:42