Difference between revisions of "2007 iTest Problems/Problem 52"

(Created page with "== Problem == Let <math>T=TNFTPP</math>. Let <math>R</math> be the region consisting of points <math>(x,y)</math> of the Cartesian plane satisfying both <math>|x|-|y|\le T-500</...")
 
(Solution to Problem 52 — drawing diagram took a lot of work)
 
Line 1: Line 1:
== Problem ==
+
''The following problem is from the Ultimate Question of the [[2007 iTest]], where solving this problem required the answer of a previous problem.  When the problem is rewritten, the T-value is substituted.''
  
Let <math>T=TNFTPP</math>. Let <math>R</math> be the region consisting of points <math>(x,y)</math> of the Cartesian plane satisfying both
+
==Problem==
<math>|x|-|y|\le T-500</math> and <math>|y|\le T-500</math>. Find the area of region <math>R</math>.
 
  
== Solution ==
+
Let <math>R</math> be the region consisting of points <math>(x,y)</math> of the Cartesian plane satisfying both
 +
<math>|x|-|y|\le 16</math> and <math>|y|\le 16</math>. Find the area of region <math>R</math>.
 +
 
 +
==Solution==
 +
 
 +
From the conditions, <math>-16 \le y \le 16</math> and <math>x \le 16 + |y|</math>.  By analyzing the conditions and testing points, we can graph the inequalities.
 +
 
 +
<asy>
 +
fill((-32,16)--(32,16)--(32,-16)--(-32,-16)--cycle,yellow);
 +
fill((-32,16)--(-16,0)--(-32,-16)--(-32,-20)--(-4,-20)--(0,-16)--(4,-20)--(32,-20)--(32,-16)--(16,0)--(32,16)--(32,20)--(4,20)--(0,16)--(-4,20)--(-32,20)--cycle,cyan);
 +
fill((-32,-16)--(32,-16)--(16,0)--(32,16)--(-32,16)--(-16,0)--cycle,green);
 +
 
 +
import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black;
 +
real xmin=-32.2,xmax=32.2,ymin=-20.2,ymax=20.2;
 +
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=2,gy=2;
 +
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=4.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=4.0,Size=2,NoZero),Arrows(6),above=true);
 +
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 +
 
 +
</asy>
 +
 
 +
The wanted area is a rectangle with two triangles cut out.  The area of the rectangle is <math>64 \cdot 32 = 2048</math>, and the area of the two triangles is <math>2 \cdot \tfrac12 \cdot 32 \cdot 16 = 512</math>.  That means the area of region <math>R</math> is <math>2048-512 = \boxed{1536}</math>.
 +
 
 +
==See Also==
 +
{{iTest box|year=2007|num-b=51|num-a=53}}
 +
 
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 16:54, 10 July 2018

The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

Problem

Let $R$ be the region consisting of points $(x,y)$ of the Cartesian plane satisfying both $|x|-|y|\le 16$ and $|y|\le 16$. Find the area of region $R$.

Solution

From the conditions, $-16 \le y \le 16$ and $x \le 16 + |y|$. By analyzing the conditions and testing points, we can graph the inequalities.

[asy] fill((-32,16)--(32,16)--(32,-16)--(-32,-16)--cycle,yellow); fill((-32,16)--(-16,0)--(-32,-16)--(-32,-20)--(-4,-20)--(0,-16)--(4,-20)--(32,-20)--(32,-16)--(16,0)--(32,16)--(32,20)--(4,20)--(0,16)--(-4,20)--(-32,20)--cycle,cyan); fill((-32,-16)--(32,-16)--(16,0)--(32,16)--(-32,16)--(-16,0)--cycle,green);  import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-32.2,xmax=32.2,ymin=-20.2,ymax=20.2;  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=2,gy=2; 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=4.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=4.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  [/asy]

The wanted area is a rectangle with two triangles cut out. The area of the rectangle is $64 \cdot 32 = 2048$, and the area of the two triangles is $2 \cdot \tfrac12 \cdot 32 \cdot 16 = 512$. That means the area of region $R$ is $2048-512 = \boxed{1536}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 51
Followed by:
Problem 53
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 51 52 53 54 55 56 57 58 59 60 TB1 TB2 TB3 TB4