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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 $$CP$$?

Aug 27, 2018

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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.

Aug 27, 2018