Difference between revisions of "1992 AJHSME Problems/Problem 22"
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==See Also== | ==See Also== | ||
{{AJHSME box|year=1992|num-b=21|num-a=23}} | {{AJHSME box|year=1992|num-b=21|num-a=23}} | ||
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Latest revision as of 23:10, 4 July 2013
Problem
Eight 
square tiles are arranged as shown so their outside edges form a polygon with a perimeter of 
units. Two additional tiles of the same size are added to the figure so that at least one side of each tile is shared with a side of one of the squares in the original figure. Which of the following could be the perimeter of the new figure?
![[asy] for (int a=1; a <= 4; ++a) { draw((a,0)--(a,2)); } draw((0,0)--(4,0)); draw((0,1)--(5,1)); draw((1,2)--(5,2)); draw((0,0)--(0,1)); draw((5,1)--(5,2)); [/asy]](http://latex.artofproblemsolving.com/0/8/a/08aaee15864e18a32aa70fb9c2f14e3dcc4a29c9.png)
Solution
One such figure would be
![[asy] for (int a=1; a <= 4; ++a) { draw((a,0)--(a,3)); } draw((0,0)--(4,0)); draw((0,1)--(5,1)); draw((1,2)--(5,2)); draw((0,0)--(0,1)); draw((5,1)--(5,2)); draw((2,3)--(1,3)); draw((4,3)--(3,3)); [/asy]](http://latex.artofproblemsolving.com/8/d/7/8d7bc4df2b6e03187fbb098512c4bbc4a4571366.png)
The perimeter of this figure is 
.
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
