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Thread: Minimisation problem

  1. #1
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    Minimisation problem

    I got stuck on this question and I hope somebody could help

    Given the triangle ABC and a point P lies inside the triangle, show that the distance AP^2 + BP^2 + CP^2 is minimum. Show further that P is the intersection point of three madians of the triangle.

    Thank you for your time

    KN
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  2. #2
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    Quote Originally Posted by knguyen2005 View Post
    Given the triangle ABC and a point P lies inside the triangle, show that the distance AP^2 + BP^2 + CP^2 is minimum. Show further that P is the intersection point of three madians of the triangle.
    I don't see an elegant geometric way to do this. Analytically, it's easy but ugly.

    Let A, B, C be the points $\displaystyle (a_1,a_2),\ (b_1,b_2),\ (c_1,c_2)$, and let $\displaystyle P=(x,y)$. Then $\displaystyle PA^2 = (x-a_1)^2 + (y-a_2)^2$, with similar expressions for $\displaystyle PB^2$ and $\displaystyle PC^2$. Add these together and you get $\displaystyle 3x^2 - 2(a_1+b_1+c_1)x + 3y^2 - 2(a_2+b_2+c_2)y + \text{const.}$ This is equal to $\displaystyle 3\bigl(x - \tfrac13(a_1+b_1+c_1)\bigr)^2 + 3\bigl(y - \tfrac13(a_2+b_2+c_2)\bigr)^2 + \text{const.}$, which is minimised when $\displaystyle x = \tfrac13(a_1+b_1+c_1)$ and $\displaystyle y = \tfrac13(a_2+b_2+c_2)$.
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