Difference between revisions of "2016 AIME I Problems/Problem 14"
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<cmath> (0,0), (1, \frac{3}{7}), (2, \frac{6}{7}), (3, 1 + \frac{2}{7}), (4, 1 + \frac{5}{7}), (5, 2 + \frac{1}{7}), (6, 2 + \frac{4}{7}), (7,3). </cmath> | <cmath> (0,0), (1, \frac{3}{7}), (2, \frac{6}{7}), (3, 1 + \frac{2}{7}), (4, 1 + \frac{5}{7}), (5, 2 + \frac{1}{7}), (6, 2 + \frac{4}{7}), (7,3). </cmath> | ||
− | We claim that the lower right vertex of the square centered at <math>(2,1)</math> lies on <math>l</math>. Since the square has side length <math>\frac{1}{5}</math>, the lower right vertex of this square has coordinates <math>(2 + \frac{1}{10}, 1 - \frac{1}{10}) = (\frac{21}{10}, \frac{9}{10})</math>. Because <math>\frac{9}{10} = \frac{3}{7} \cdot \frac{21}{10}</math>, <math>(\frac{21}{10}, \frac{9}{10})</math> lies on <math>l</math>. Since the circle centered at <math>(2,1)</math> is contained inside the square, this circle does not intersect <math>l</math>. Similarly the upper left vertex of the square centered at <math>(5,2)</math> is on <math>l</math>. Since every other point listed above is farther away from a lattice point (excluding (0,0) and (7,3)) and there are two squares with centers strictly between <math>(0,0)</math> and <math>(7,3)</math> that intersect <math>l</math>. Since there are <math>\frac{1001}{7} = \frac{429}{3} = 143</math> segments from <math>(7k, 3k)</math> to <math>(7(k + 1), 3(k + 1))</math>, the above count is yields <math>143 \cdot 2 = 286</math> | + | We claim that the lower right vertex of the square centered at <math>(2,1)</math> lies on <math>l</math>. Since the square has side length <math>\frac{1}{5}</math>, the lower right vertex of this square has coordinates <math>(2 + \frac{1}{10}, 1 - \frac{1}{10}) = (\frac{21}{10}, \frac{9}{10})</math>. Because <math>\frac{9}{10} = \frac{3}{7} \cdot \frac{21}{10}</math>, <math>(\frac{21}{10}, \frac{9}{10})</math> lies on <math>l</math>. Since the circle centered at <math>(2,1)</math> is contained inside the square, this circle does not intersect <math>l</math>. Similarly the upper left vertex of the square centered at <math>(5,2)</math> is on <math>l</math>. Since every other point listed above is farther away from a lattice point (excluding (0,0) and (7,3)) and there are two squares with centers strictly between <math>(0,0)</math> and <math>(7,3)</math> that intersect <math>l</math>. Since there are <math>\frac{1001}{7} = \frac{429}{3} = 143</math> segments from <math>(7k, 3k)</math> to <math>(7(k + 1), 3(k + 1))</math>, the above count is yields <math>143 \cdot 2 = 286</math> squares. Since every lattice point on <math>l</math> is of the form <math>(3k, 7k)</math> where <math>0 \le k \le 143</math>, there are <math>144</math> lattice points on <math>l</math>. Centered at each lattice point, there is one square and one circle, hence this counts <math>288</math> squares and circles. Thus <math>m + n = 286 + 288 = \boxed{574}</math>. |
(Solution by gundraja) | (Solution by gundraja) | ||
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for(int i=0;i<8;++i)for(int j=0;j<4;++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));} | for(int i=0;i<8;++i)for(int j=0;j<4;++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));} | ||
</asy> | </asy> | ||
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== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=13|num-a=15}} | {{AIME box|year=2016|n=I|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:17, 5 March 2016
Problem
Centered at each lattice point in the coordinate plane are a circle radius and a square with sides of length whose sides are parallel to the coordinate axes. The line segment from to intersects of the squares and of the circles. Find .
Solution
First note that and so every point of the form is on the line. Then consider the line from to . Translate the line so that is now the origin. There is one square and one circle that intersect the line around . Then the points on with an integral -coordinate are, since has the equation :
We claim that the lower right vertex of the square centered at lies on . Since the square has side length , the lower right vertex of this square has coordinates . Because , lies on . Since the circle centered at is contained inside the square, this circle does not intersect . Similarly the upper left vertex of the square centered at is on . Since every other point listed above is farther away from a lattice point (excluding (0,0) and (7,3)) and there are two squares with centers strictly between and that intersect . Since there are segments from to , the above count is yields squares. Since every lattice point on is of the form where , there are lattice points on . Centered at each lattice point, there is one square and one circle, hence this counts squares and circles. Thus .
(Solution by gundraja)
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.