Difference between revisions of "1992 AJHSME Problems/Problem 22"

(Created page with "==Problem== Eight <math>1\times 1</math> square tiles are arranged as shown so their outside edges form a polygon with a perimeter of <math>14</math> units. Two additional tile...")
 
 
(One intermediate revision by one other user not shown)
Line 17: Line 17:
  
 
<math>\text{(A)}\ 15 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20</math>
 
<math>\text{(A)}\ 15 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20</math>
 +
 +
==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>
 +
 +
The perimeter of this figure is <math>\boxed{\text{(C)}\ 18}</math>.
 +
 +
==See Also==
 +
{{AJHSME box|year=1992|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Latest revision as of 23:10, 4 July 2013

Problem

Eight $1\times 1$ square tiles are arranged as shown so their outside edges form a polygon with a perimeter of $14$ 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]


$\text{(A)}\ 15 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$

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]

The perimeter of this figure is $\boxed{\text{(C)}\ 18}$.

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
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. AMC logo.png