Difference between revisions of "1993 AJHSME Problems/Problem 25"
Line 6: | Line 6: | ||
==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> | ||
+ | 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== | ==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, inches on a side, is placed on the board so that it covers part or all of the area of each of squares. The maximum possible value of is
Solution
Using Pythagorean Theorem, the diagonal of the square . Because this is longer than , the length of the sides of two adjacent squares, the card can be placed like so, covering squares. .
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 |