We have a right triangle \(\triangle ABC\) where the legs \(AB\) and \(BC\) have lengths \(6\) and \(3\sqrt{3}\) respectively. Medians \(AM\) and \(CN\) meet at point \(P.\) What is the length of \(CP\)?

Guest Aug 27, 2018

#1**+1 **

Put it on a coordinate plane.

Now A(0,6), B(0,0), C(\(3\sqrt3\),0)

Slope of AM = \(\dfrac{0-6}{\dfrac{3\sqrt3}{2}-0}\)=\(-\dfrac{4\sqrt3}{3}\)

Equation of AM is \(y = -\dfrac{4\sqrt3x}{3}+6\)

Slope of CN = \(\dfrac{0-3}{3\sqrt3-0}\)=\(-\dfrac{\sqrt3}{3}\)

Equation of CN is \(y=-\dfrac{\sqrt3x}{3}+3\)

Solving for the intersection point(P):

P = (\(\sqrt3\),2)

Recalling that C = (\(3\sqrt3\),0).

Length of CP = \(\sqrt{(\sqrt3-0)^2+(3\sqrt3-2)^2}\) = \(\sqrt{34-12\sqrt3}\).

And this is the simplest it can get.

MaxWong Aug 27, 2018