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Given a right triangle with sides a,b and a hypotenuse c=2, calculate the sum of trianle's squared medians i.e. if medians are x,y, and z, calculate

$x^2+y^2+z^2$

The only thing i thought of is using the property that medians devide each other in 2:1 ratio.

Thanks ;)

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  • $\begingroup$ Did you draw a picture? $\endgroup$ – 5xum Jun 30 '14 at 11:36
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Take a look at this: http://en.wikipedia.org/wiki/Right_triangle#Medians

By drawing a picture:

http://www.geometryexpressions.com/explore.php?p=04-Examples%2F02-Example_Book%2F01-Triangles&f=Example+007-Intersection_Of_The_Medians_Of_A_Right_Angled_Triangle.htm

It is derived that

$$x^2 + y^2 + z^2 = \frac{3}{4}*(a^2 + b^2 + c^2)$$

Since $a^2 + b^2 = c^2$, we have that $$x^2 + y^2 + z^2 = \frac{3}{4}*(2c^2)$$

Therefore $$x^2 + y^2 + z^2 = \frac{3}{4}*(2*(2^2))$$ $$x^2 + y^2 + z^2 = 6$$

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  • $\begingroup$ Oh my. If only i remembered that the angle above the arc is right.. :( Thanks :D $\endgroup$ – Transcendental Jun 30 '14 at 12:32
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consider that the triangle is on a grind constructed so that the side $a$ is on the $(Ox)$ axis and the side $b$ on the $(Oy)$ axis.

You can compute the coordinates of each points you need, and all the distances you like.

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