Difference between revisions of "1976 AHSME Problems/Problem 8"
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− | In order for the coordinates to have absolute values less than <math>4</math>, the points must lie in the <math>8</math> by <math>8</math> square passing through the points <math>(-4,-4), (-4, 4), (4,4)</math> and <math>(4, -4)</math>. | + | In order for the coordinates to have absolute values less than <math>4</math>, the points must lie in or on the <math>8</math> by <math>8</math> square passing through the points <math>(-4,-4), (-4, 4), (4,4)</math> and <math>(4, -4)</math>. There are <math>(8+1)(8+1) = 81</math> points in or on this region. |
− | + | Since the point must have integer coordinates, we have to count the number of points within or on <math>x^2 + y^2 = 2^2</math>. By graphing and counting, there are 13 options. | |
− | < | + | <math>\boxed{A \frac{13}{81}</math> |
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− | ~ | + | ~thshea |
Revision as of 12:10, 29 June 2022
Problem 8
A point in the plane, both of whose rectangular coordinates are integers with absolute values less than or equal to four, is chosen at random, with all such points having an equal probability of being chosen. What is the probability that the distance from the point to the origin is at most two units?
Solution
In order for the coordinates to have absolute values less than , the points must lie in or on the by square passing through the points and . There are points in or on this region. Since the point must have integer coordinates, we have to count the number of points within or on . By graphing and counting, there are 13 options. $\boxed{A \frac{13}{81}$ (Error compiling LaTeX. Unknown error_msg)
~thshea