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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 Sharma
3 weeks, 4 days 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 weeks ago

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@Ajinkya Shivashankar Yep, That's a great observation. Aditya Sharma · 2 weeks, 6 days ago

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@Aditya Sharma What was ur solution? Harsh Shrivastava · 2 weeks, 6 days 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 Sharma · 2 weeks, 6 days ago

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@Ajinkya Shivashankar Awesome solution! Harsh Shrivastava · 3 weeks ago

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