1982 AHSME Problems/Problem 9
Problem
A vertical line divides the triangle with vertices 
, and 
in the 
into two regions of equal area.
The equation of the line is 
Solution
![[asy] size(350); defaultpen(fontsize(10)); pair A=origin, O=(10,0), B=(3,0), N=(0,5), C=(3,5), P=(5,0), D=(1,1), G=(9,1), F=(1,0); draw(G--A--D--cycle, linewidth(0.7)); draw(D--F, linewidth(0.6)); draw(B--C, linewidth(0.5)); draw(A--N, linewidth(0.4)); draw(A--O, linewidth(0.4)); dot(A^^B^^D^^G^^F); label("$A$", D, W); label("$B$", A, dir(0)); label("$C$", G, dir(100)); label("$E$", B, SE); label("$F$", F, SE); label("$(9,1)$", G, SE); label("$(0,0)$", A, SW); label("$8$", D--G, dir(40)); label("$y=x/9$", A--G, SE); label("$x=a$", B--N, SE); label("$(1,1)$", D, NE);[/asy]](http://latex.artofproblemsolving.com/1/6/1/161d96fc6e6ee3d71b1828965b66eafd057660e5.png)
The vertical line that divides 
into two equal regions has equation 
, as shown in the diagram.
The area of 
is 
, so the two regions must each have area 
.
Since 
has area 
, the portion of to the left of 
will be less than and therefore less than . So 
is to the left of vertical line (Passing through point 
).
The equation of line BC is 
and the vertical line intersects 
at the point 
. Because the area of the portion of on the right is , we have ![\[2=\frac{1}{2}\left(1-\frac{a}{9}\right)(9-a)\]](http://latex.artofproblemsolving.com/b/e/c/bec39533658c91ff4fa77eb9669c60e1d4c01cc4.png)
which simplifies to ![\[(9-a)^2=36\]](http://latex.artofproblemsolving.com/e/0/5/e05e9429ce5f454c880cbe90f48d10c252ed96a6.png)
This equation has solutions 
and 
, but 
so 
.
See Also
| 1982 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
