1986 AHSME Problems/Problem 11
Problem
In 
and 
. Also, 
is the midpoint of side 
and 
is the foot of the altitude from 
to 
.
The length of 
is
![[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A); draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C)); label("A", A, NE); label("B", B, W); label("C", C, E); label("H", H, S); label("M", M, dir(M)); [/asy]](http://latex.artofproblemsolving.com/4/3/9/439be0c9c9fcdca68ec7ff37d19968d3ec0f7c14.png)
Solution
In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is 
, the median must be 
.
Solution 2 (Self Torture)
Warning: this solution is very intensive in calculation. Please do NOT try this on the test!
Let's start by finding 
. By Heron's Formula, ![$s=\frac{13+14+15}{2}=21, [ABC]=\sqrt{21*(21-13)(21-14)(21-15)}=84$](http://latex.artofproblemsolving.com/1/0/1/101c89420dd448e6204babea290582b1b90b0018.png)
. Using the area formula 
, 
. Now using the Pythagorean Theorem, 
.
Now 
. Using Stewart's Theorem on 
, letting 
:
(remember that Stewart's Theorem is 
).
Thus 
or 
(reject this solution since 
is positive). Thus 
. Select 
.
~hastapasta
P.S.: Although this is torturous, this is a good practice of Heron's formula and Stewart's theorem though.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
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