Difference between revisions of "1993 AJHSME Problems/Problem 25"

Line 23: Line 23:
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1993|num-b=24|after=Last <br /> Problem}}
 
{{AJHSME box|year=1993|num-b=24|after=Last <br /> Problem}}
 +
{{MAA Notice}}

Revision as of 00:12, 5 July 2013

Problem

A checkerboard consists of one-inch squares. A square card, $1.5$ inches on a side, is placed on the board so that it covers part or all of the area of each of $n$ squares. The maximum possible value of $n$ is

$\text{(A)}\ 4\text{ or }5 \qquad \text{(B)}\ 6\text{ or }7\qquad \text{(C)}\ 8\text{ or }9 \qquad \text{(D)}\ 10\text{ or }11 \qquad \text{(E)}\ 12\text{ or more}$

Solution

Using Pythagorean Theorem, the diagonal of the square $\sqrt{(1.5)^2+(1.5)^2}=\sqrt{4.5}>2$. Because this is longer than $2$, the length of the sides of two adjacent squares, the card can be placed like so, covering $12$ squares. $\rightarrow \boxed{\text{(E)}\ 12\ \text{or more}}$.

[asy] for (int a = -2; a <= 2; ++a) {     draw((-2,a)--(2,a)); draw((a,-2)--(a,2)); } pair A,B,C,D; A=(0,sqrt(2.25)); B=(sqrt(2.25),0); C=(0,-sqrt(2.25)); D=(-sqrt(2.25),0); draw(A--B--C--D--cycle); fill(A--B--C--D--cycle,lightgray); [/asy]


See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last
Problem
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