Difference between revisions of "1989 AHSME Problems/Problem 5"
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in which case the calculation is <math>2(20\cdot 10)+20+10=\boxed{430}</math>. | in which case the calculation is <math>2(20\cdot 10)+20+10=\boxed{430}</math>. | ||
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| + | == See also == | ||
| + | {{AHSME box|year=1989|num-b=4|num-a=6}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 06:43, 22 October 2014
Problem
Toothpicks of equal length are used to build a rectangular grid as shown. If the grid is 20 toothpicks high and 10 toothpicks wide, then the number of toothpicks used is
![[asy] real xscl = 1.2; int[] x = {0,1,2,4,5},y={0,1,3,4,5}; for(int a:x){ for(int b:y) { dot((a*xscl,b)); } } for(int a:x) { pair prev = (a,y[0]); for(int i = 1;i<y.length;++i) { pair p = (a,y[i]); pen pen = linewidth(.7); if(y[i]-prev.y!=1){ pen+=dotted; } draw((xscl*prev.x,prev.y)--(xscl*p.x,p.y),pen); prev = p; } }for(int a:y) { pair prev = (x[0],a); for(int i = 1;i<x.length;++i) { pair p = (x[i],a); pen pen = linewidth(.7); if(x[i]-prev.x!=1){ pen+=dotted; } draw((xscl*prev.x,prev.y)--(p.x*xscl,p.y),pen); prev = p; } } path lblx = (0,-.7)--(5*xscl,-.7); draw(lblx); label("$10$",lblx); path lbly = (5*xscl+.7,0)--(5*xscl+.7,5); draw(lbly); label("$20$",lbly);[/asy]](http://latex.artofproblemsolving.com/a/2/c/a2c4cc06b98e5b80f7a0bd3b5d8e48991f7a8e55.png)
Solution
There are 21 horizontal lines made of 10 matches, and 11 vertical lines made of 20 matches, and 
.
Alternatively, the frame can be dissected into 
L shapes, the right-hand border, and the bottom border:
![[asy] path p,q; for(int i=0;i<3;++i) { for(int j=3;j<6;++j) { p = (i+0.75,j)--(i,j)--(i,j-0.75); draw(p); dot((i,j)); } } for(int i=0;i<3;++i) { q = (i,0)--(i+0.75,0); draw(q);dot((i,0)); q = (5,5-i)--(5,4.25-i); draw(q);dot((5,5-i)); } draw((4,0)--(4.75,0));dot((4,0)); draw((5,1)--(5,0.25));dot((5,1)); draw((3,2)--(4,1),dotted); [/asy]](http://latex.artofproblemsolving.com/f/1/8/f1851f7805bf60317c9acb84ce33d853432f75fe.png)
in which case the calculation is 
.
See also
| 1989 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
