1961 AHSME Problems/Problem 20
Problem
The set of points satisfying the pair of inequalities 
and 
is contained entirely in quadrants:
Solution
![[asy] fill((-1,5)--(5,5)--(5,-1)--cycle,red); fill((2.5,5)--(-5,5)--(-5,-5)--(-2.5,-5)--cycle,cyan); fill((-1,5)--(4/3,8/3)--(2.5,5)--cycle,magenta); draw((-1,5)--(5,-1),dotted,Arrows); draw((-2.5,-5)--(2.5,5),dotted,Arrows); import graph; size(8.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=5.2,ymin=-5.2,ymax=5.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](http://latex.artofproblemsolving.com/6/d/6/6d6f3e0047318d2940e25e439a2ce5999da20deb.png)
Graph both equations. The solution to the system of inequalities is the magenta area because it is in both of the solutions of each individual inequality. From the graph, the magenta area is in quadrants 
and 
, so the answer is 
.
See Also
| 1961 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
