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A geometric inequality

In a \(\Delta\)ABC if \(P\) is any interior point inside it, Find the minimum value of :

\(\displaystyle \text{min}(BC.PA^2+CA.PB^2+AB.PC^2)\)

Note by Aditya Narayan Sharma
3 months, 2 weeks ago

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keep 3 masses of values BC,CA,AB at points A,B,C. Moment of inertia about any point P is given by (BC.PA^2+...)which is what we have to minimize. However moment of inertia is minimum at centre of mass so we find centre of mass whose coordinates are Same as formula of coordinates of incentre given sides. So the required point P is its incentre Ajinkya Shivashankar · 3 months, 1 week ago

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@Ajinkya Shivashankar Yep, That's a great observation. Aditya Narayan Sharma · 3 months, 1 week ago

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@Aditya Narayan Sharma What was ur solution? Harsh Shrivastava · 3 months, 1 week ago

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@Harsh Shrivastava That used the Erdos-Mordell inequality and then some co-ordinate, anyways his observation from physics outweighs mine here. Aditya Narayan Sharma · 3 months, 1 week ago

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@Ajinkya Shivashankar Awesome solution! Harsh Shrivastava · 3 months, 1 week ago

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