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 Equidistant point (Posted on 2018-06-03)
Given a triangle ABC and a straight line L.

Find the point P on L such that
PA2+PB2+PC2 is the smallest.

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 one messy method | Comment 1 of 4
If we call L the x-axis we can arbitrarily represent any triangle as three points above this axis, at heights y1, y2, and y3. Let us put the first point on the y-axis (x1=0) and place the other two at x2 and x3 so the three points of the triangle are (0,y1) (x2,y2), (x3,y3). Placing point P on the x-axis at x, that is (x,0). Note that this arrangement maintains the generality of the problem. The sum of the three lengths squared as drawn from (x,0) are:

sum = y1^2 x^2 + (x2-x)^2 y2^2 + (x3-x)^2 y3^2

This sum is minimum when d(sum)/dx = 0

This occurs when

x = (x2 y2^2 + x3 y3^2) / (y1^2 + y2^2 + y3^2)

While all this may be true - I bet there is some elegant construction drawn from the triangle that simply intersects at P.

 Posted by Steven Lord on 2018-06-04 23:48:05

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