Difference between revisions of "2016 AIME I Problems/Problem 14"
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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> circles. 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>. | 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> circles. 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) |
| − | |||
== 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 23:23, 4 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 
:
![\[(0,0), (1, \frac{3}{7}), (2, \frac{6}{7}), (3, 1 + \frac{2}{7}), (4, 1 + \frac{5}{7}), (5, 2 + \frac{1}{7}), (6, 2 + \frac{4}{7}), (7,3).\]](http://latex.artofproblemsolving.com/8/9/1/891edbf0029fe08e770b58312fe80a4e1f1f0371.png)
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 
circles. 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. 
