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
1 year, 7 months 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 - 1 year, 7 months ago

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Yep, That's a great observation.

Aditya Narayan Sharma - 1 year, 7 months ago

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What was ur solution?

Harsh Shrivastava - 1 year, 7 months 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 - 1 year, 7 months ago

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Awesome solution!

Harsh Shrivastava - 1 year, 7 months ago

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