+238 In right triangle $ABC,$ $\angle C = 90^\circ$. Median $\overline{AM}$ has a length of 1, and median $\overline{BN}$ has a length of 1. What is the length of the hypotenuse of the triangle?
+1908 Let \(BC = a, AC = b\) just to simplify our eyes a bit.
Let's note that through the Pythagorean Theorem, we can find AM and BN in terms of a and b.
We get \(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)
and \(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)
Because AM and BN both equal 1, we get the equations
\(4b^2 + a^2 = 1^2 \\4a^2 + b^2 = 1^2\)
Adding the two equations together and dividing, we get
\(a^2 + b^2 = \dfrac{1+ 1}{5} = 2/5\)
Since the hypotenuse is just the square root of a^2 + b^2, we get
\(\sqrt{a^2+ b^2} = \sqrt{2/5}\)
Thanks! :)
