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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > Let P be an interior point of an equilateral triangle...
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Let P be an interior point of an equilateral triangle ABC, such that AP^2 = BP^2 + CP^2, then find the value of 8{sin(

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The angle BPC is double of angle A. We know the angle at A is 60 degrees (since equilateral) So < BPC = 120 degrees 8 { sin (
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Without loss of generality may assume that the sides of the triangle have length 1. May choose the coordinate axes so that A, B and C have coordinates (0,0), (1,0) and ( 1/2,root(3)/2 ). Let (x,y) be the coordinates of the point P. Then x^2+y^2 = (x-1)^2+y^2+(x-1/2)^2+(y-root(3)/2)^2 . i.e. x^2-3x+5/4+(y-root(3)/2)^2...
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Without loss of generality may assume that the sides of the triangle have length 1. May choose the coordinate axes so that A, B and C have coordinates (0,0), (1,0) and ( 1/2,root(3)/2 ). Let (x,y) be the coordinates of the point P. Then x^2+y^2 = (x-1)^2+y^2+(x-1/2)^2+(y-root(3)/2)^2 . i.e. x^2-3x+5/4+(y-root(3)/2)^2 = 0. or (x-3/2)^2+(y-root(3)/2)^2 = 1. Thus the locus of the point P is a circle of radius 1 with the center at the point D with coordinates (3/2,root(3)/2). Clearly ABDC is a parallelogram and so the read less
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Without loss of generality may assume that the sides of the triangle have length 1. May choose the coordinate axes so that A, B and C have coordinates (0,0), (1,0) and ( 1/2,root(3)/2 ). Let (x,y) be the coordinates of the point P. Then x^2+y^2 = (x-1)^2+y^2+(x-1/2)^2+(y-root(3)/2)^2 . i.e. x^2-3x+5/4+(y-root(3)/2)^2...
read more
Without loss of generality may assume that the sides of the triangle have length 1. May choose the coordinate axes so that A, B and C have coordinates (0,0), (1,0) and ( 1/2,root(3)/2 ). Let (x,y) be the coordinates of the point P. Then x^2+y^2 = (x-1)^2+y^2+(x-1/2)^2+(y-root(3)/2)^2 . i.e. x^2-3x+5/4+(y-root(3)/2)^2 = 0. or (x-3/2)^2+(y-root(3)/2)^2 = 1. Thus the locus of the point P is a circle of radious 1 with the centre at the point D with coordinates (3/2,root(3)/2). It passes through B and C. Clearly ABDC is a parallelogram and so the read less
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