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>.
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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>( 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>
 
<asy>
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fill(A--B--C--D--cycle,lightgray);
 
fill(A--B--C--D--cycle,lightgray);
 
</asy>
 
</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}}
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{{MAA Notice}}

Latest revision as of 22:07, 25 May 2021

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 $2$( 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

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