1975 AHSME Problems/Problem 20
Revision as of 16:29, 19 January 2021 by Hashtagmath (talk | contribs)
Problem
In the adjoining figure triangle 
is such that 
and 
. IF 
is the midpoint of 
and 
, what is the length of ?
![[asy] draw((-4,0)--(4,0)--(-1,4)--cycle); draw((-1, 4)--(0, 0.00001)); label("B", (-4,0), S); label("C", (4,0), S); label("A", (-1, 4), N); label("M", (0, 0.0001), S); [/asy]](http://latex.artofproblemsolving.com/f/4/2/f42fe5450102486dda1fdec4c1520e55ac6dee3e.png)
Solution
Let 
. Then, by Stewart's Theorem, we have
![\[2x^3+18x=16x+64x\]](http://latex.artofproblemsolving.com/a/f/d/afd3cf1a1a368ef79905d4e8d3fa41029f467bc9.png)
![\[\implies x^2+9=40\]](http://latex.artofproblemsolving.com/e/3/9/e391428c480e31e2bb9e9ff00e008e3893142584.png)
![\[\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.\]](http://latex.artofproblemsolving.com/1/8/1/181584bcc40552dafcf9b933095297d4ac21c89c.png)
The answer is 
-brainiacmaniac31
See Also
| 1975 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
