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There is a triangle $ABC$ where $|CB|=a$, $|AC|=b$ and medians of these sides intersect at a right angle. Find |AB|.

I don't know how to use a right angle in this problem. I have a idea to link a middle of $|CB|$ with $|AC|$, let $K,L$ be a centre of these sides and we have $2|KL|=|AB|$

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We've the known medians theorem in geometry: the three medians to the sides of a triangle meet at one point which divides each median in a ratio of $\;1:2\;$ , with the longest segment always on the vertex side.

Thus, calling $\;M\;$ to the intersection point of the two medians, we can put

$$|AM|=2x\;,\;\;|MD|=x\;\;;\;\;|BM|=2y\;,\;\;|ME|=y$$

with $\;D=\;$ the midpoint of $\;BC\;$ , and $\;E=\;$ the midpoint of AC.

Using now Pythagoras on the triangle $\;\Delta AMB\;$ we get:

$$|AM|^2+|BM|^2=|AB|^2=4(x^2+y^2)$$

But on $\;\Delta BMD\;$ we get:

$$4y^2+x^2=\frac{a^2}4$$

and on $\;\Delta AME\;$ we get

$$4x^2+y^2=\frac{b^2}4$$

so solving the quadratic system of two unknowns above, we get:

$$|AB|=2\sqrt{x^2+y^2}=2\sqrt{\frac{4b^2-a^2}{60}+\frac{4a^2-b^2}{60}}=\ldots$$

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