1951 AHSME Problems/Problem 49
The medians of a right triangle which are drawn from the vertices of the acute angles are and . The value of the hypotenuse is:
We will proceed by coordinate bashing.
Call the first leg , and the second leg (We are using the double of a variable to avoid any fractions)
Notice that we want to find
Two equations can be written for the two medians: and .
Add them together and we get ,
Dividing by 5 gives
Multiply them by 4 gives , just what we need to find the hypotenuse. Recall that he hypotenuse is . The value inside the radical is equal to , so the hypotenuse is equal to