1958 AHSME Problems/Problem 43
Problem
is the hypotenuse of a right triangle 
. Median 
has length 
and median 
has length 
. The length of is:
Solution
![[asy] import geometry; unitsize(50); pair A = (0,0), B = (3,0), C = (0, 4); pair AB = midpoint(A--B), AC = midpoint(A--C); draw(A--B--C--A); draw(A--B, StickIntervalMarker(2, 1)); draw(A--C, StickIntervalMarker(2, 2)); draw(C--AB); draw(B--AC); dot(AB); dot(AC); MP("$A$", A, W); MP("$B$", B, E); MP("$C$", C, W); MP("$M$", AB, S); MP("$N$", AC, W); label("$x$", A--AB, S); label("$x$", AB--B, S); label("$y$", A--AC, W); label("$y$", AC--C, W); draw(rightanglemark(C, A, B)); [/asy]](http://latex.artofproblemsolving.com/c/e/2/ce21d644f5a51b97c81a5a0c36707ab331712567.png)
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 42 |
Followed by Problem 44 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
