Difference between revisions of "2010 AIME I Problems/Problem 8"
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== Solution == | == Solution == | ||
− | The desired region consists of 12 boxes, whose lower-left corners are integers solutions of <math>x^2 + y^2 = 25</math>, namely <math>(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).</math> Since the points themselves are symmetric about <math>(0,0)</math>, the boxes are symmetric about <math>\left(\frac12,\frac12\right)</math>. The distance from <math>\left(\frac12,\frac12\right)</math> to the furthest point on an axis | + | The desired region consists of 12 boxes, whose lower-left corners are integers solutions of <math>x^2 + y^2 = 25</math>, namely <math>(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).</math> Since the points themselves are symmetric about <math>(0,0)</math>, the boxes are symmetric about <math>\left(\frac12,\frac12\right)</math>. The distance from <math>\left(\frac12,\frac12\right)</math> to the furthest point on a box that lays on an axis, for instance <math>(6,1)</math>, is <math>\sqrt {\frac {11}2^2 + \frac12^2} = \sqrt {\frac {122}4}.</math> The distance from <math>\left(\frac12,\frac12\right)</math> to the furthest point on a box in the middle of a quadrant, for instance <math>(5,4)</math>, is <math>\sqrt {\frac92^2 + \frac72^2} = \sqrt {\frac {130}4}.</math> The latter is the larger, and is <math>\frac {\sqrt {130}}2</math>, giving an answer of <math>130 + 2 = \boxed{132}</math>. |
<center><asy>import graph; size(10.22cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.68,xmax=6.54,ymin=-5.52,ymax=6.5; | <center><asy>import graph; size(10.22cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.68,xmax=6.54,ymin=-5.52,ymax=6.5; |
Revision as of 23:16, 5 March 2020
Problem
For a real number , let denominate the greatest integer less than or equal to . Let denote the region in the coordinate plane consisting of points such that . The region is completely contained in a disk of radius (a disk is the union of a circle and its interior). The minimum value of can be written as , where and are integers and is not divisible by the square of any prime. Find .
Solution
The desired region consists of 12 boxes, whose lower-left corners are integers solutions of , namely Since the points themselves are symmetric about , the boxes are symmetric about . The distance from to the furthest point on a box that lays on an axis, for instance , is The distance from to the furthest point on a box in the middle of a quadrant, for instance , is The latter is the larger, and is , giving an answer of .
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.