Difference between revisions of "1972 AHSME Problems/Problem 28"

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==Solution==
 
==Solution==
Consider the upper right half of the grid, which consists of a 4x4 section of the checkerboard and a quarter-circle of radius 4. We can draw this as a coordinate grid and shade in the complete squares. There are 8 squares in the upper right corner, so there are <math>8 cdot 4 = \boxed{32}</math> whole squares in total.
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Consider the upper right half of the grid, which consists of a 4x4 section of the checkerboard and a quarter-circle of radius 4. We can draw this as a coordinate grid and shade in the complete squares. There are 8 squares in the upper right corner, so there are <math>8 \cdot 4 = \boxed{32}</math> whole squares in total.
  
 
The answer is <math>\textbf{(E)}.</math>
 
The answer is <math>\textbf{(E)}.</math>
  
 
-edited by coolmath34
 
-edited by coolmath34

Revision as of 13:53, 23 June 2021

Problem 28

A circular disc with diameter $D$ is placed on an $8\times 8$ checkerboard with width $D$ so that the centers coincide. The number of checkerboard squares which are completely covered by the disc is

$\textbf{(A) }48\qquad \textbf{(B) }44\qquad \textbf{(C) }40\qquad \textbf{(D) }36\qquad  \textbf{(E) }32$

Solution

Consider the upper right half of the grid, which consists of a 4x4 section of the checkerboard and a quarter-circle of radius 4. We can draw this as a coordinate grid and shade in the complete squares. There are 8 squares in the upper right corner, so there are $8 \cdot 4 = \boxed{32}$ whole squares in total.

The answer is $\textbf{(E)}.$

-edited by coolmath34