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Difference between revisions of "1993 AJHSME Problems/Problem 25"
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==Solution==
 
==Solution==
 +
Using [[Pythagorean Theorem]], the diagonal of the square <math>\sqrt{(1.5)^2+(1.5)^2}=\sqrt{4.5}>2</math>. Because this is longer than <math>2</math>, the length of the sides of two adjacent squares, the card can be placed like so, covering <math>12</math> squares. <math>\rightarrow \boxed{\text{(E)}\ 12\ \text{or more}}</math>.
 +
 +
<asy>
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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);
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fill(A--B--C--D--cycle,lightgray);
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</asy>
  
  
 
==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}}

Revision as of 23:05, 22 December 2012

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 (Problems • Answer Key • Resources)
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
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