ABC is a triangle and 'P' is the point in the Plane of Triangle ABC. If PA^2 sinA + PB^2 sinB + PC^2 sinC is minimum then prove that P is Incentre
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I'll use vector notation, so I'm treating A,B,CABC, and PP as vectors.
By a translation, we may assume A+B−C=0ABC0, and that will make computations easier.
We're to find PP such that
∥P−A∥2+∥P−B∥2=∥P−C∥2PA2PB2PC2
The length of a vector squared can be written as a dot product of it with itself: ∥X∥2=X⋅XX2XX. So we can write the condition as
(P−A)⋅(P−A)+(P−B)⋅(P−B)=(P−C)⋅(P−C)PAPAPBPBPCPC
which can be simplified
∥P∥2−2A⋅P−∥A∥2+∥P∥2−2B⋅P−∥B∥2=∥P∥2−2C⋅P−∥C∥2P22APA2P22BPB2P22CPC2
∥P∥2+2(C−A−B)⋅P=∥A∥2+∥B∥2−∥C∥2P22CABPA2B2C2
The term 2(C−A−B)⋅P2CABP is 00 since A+B−C=0ABC0, so the equation simplified to
∥P∥2=dP2d
where dd is the constant on the right. That's the equation of a circle if we're in the plane centered at the origin. If the three points are in 3-space, then the locus is a sphere centered at the origin.
By a translation, we may assume A+B−C=0ABC0, and that will make computations easier.
We're to find PP such that
∥P−A∥2+∥P−B∥2=∥P−C∥2PA2PB2PC2
The length of a vector squared can be written as a dot product of it with itself: ∥X∥2=X⋅XX2XX. So we can write the condition as
(P−A)⋅(P−A)+(P−B)⋅(P−B)=(P−C)⋅(P−C)PAPAPBPBPCPC
which can be simplified
∥P∥2−2A⋅P−∥A∥2+∥P∥2−2B⋅P−∥B∥2=∥P∥2−2C⋅P−∥C∥2P22APA2P22BPB2P22CPC2
∥P∥2+2(C−A−B)⋅P=∥A∥2+∥B∥2−∥C∥2P22CABPA2B2C2
The term 2(C−A−B)⋅P2CABP is 00 since A+B−C=0ABC0, so the equation simplified to
∥P∥2=dP2d
where dd is the constant on the right. That's the equation of a circle if we're in the plane centered at the origin. If the three points are in 3-space, then the locus is a sphere centered at the origin.
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