Difference between revisions of "1958 AHSME Problems/Problem 43"
Treetor10145 (talk | contribs) (→Solution) |
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label("$x$", A--AB, S); | label("$x$", A--AB, S); | ||
label("$x$", AB--B, S); | label("$x$", AB--B, S); | ||
| − | label("$y$", A--AC, | + | label("$y$", A--AC, W); |
label("$y$", AC--C, W); | label("$y$", AC--C, W); | ||
draw(rightanglemark(C, A, B)); | draw(rightanglemark(C, A, B)); | ||
</asy> | </asy> | ||
| + | |||
| + | |||
| + | |||
<math>\fbox{D}</math> | <math>\fbox{D}</math> | ||
Revision as of 11:32, 23 February 2018
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
