+5
Suppose that there is a right triangle \(ABC\) such that the legs \(AB\) and \(AC\) have lengths \(4\sqrt2\) and \(2\) respectively. What is the length of the median \(BM\)?
This question is posted here by @CPhill: https://web2.0calc.com/questions/suppose-that-there-is-a-right-triangle-abc-such-that, but that answer is wrong.
Here is the correct answer:
Median \(BM \) splits side \(AC \) into two segments with equal length. Since the midpoint of the hypotenuse of a right triangle is also the circumcenter of that triangle, \(AM\) and \(BM \) are circumradii of triangle \(ABC\). Moreover, since the circumcircle of a triangle intersects each corner point of a triangle, \(BM\) is also a circumradius. Thus, \(AM = MC = BM\).
We can easily find the length of \(BM \) now. By the Pythagorean theorem,
\((4\sqrt2)^2 + 2^2 = AC^2, \)
\(32 + 4 = AC^2, \)
\(\sqrt36 = AC, \)
\(AC = 6.\)
Since \(BM = AM = MC\), \(AC = 2BM\). And since \(AC = 6\), we have
\(\boxed{BM=3}\)
\(Q.E.D.\)
