Difference between revisions of "2016 AMC 12B Problems/Problem 11"
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==Solution== | ==Solution== | ||
Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
+ | Revised by Kinglogic and RJ5303707 | ||
+ | <asy> | ||
+ | Label l; | ||
+ | l.p=fontsize(8); | ||
+ | xaxis(-1,8,Ticks(l, 1.0)); | ||
+ | yaxis(-1,16,Ticks(l, 1.0)); | ||
+ | real f(real x) | ||
+ | { | ||
+ | return x * pi; | ||
+ | } | ||
+ | D(graph(f,-1/pi,5.1)); | ||
+ | D((5.1,-1)--(5.1,16)); | ||
+ | D((-1,-0.1)--(8,-0.1)); | ||
+ | for(int x = 0; x < 5.1; ++x) { | ||
+ | for(int y = 0; y < 16; ++y) { | ||
+ | if(x * pi > y) { | ||
+ | D((x,y)); | ||
+ | } | ||
+ | } | ||
+ | } | ||
+ | </asy> | ||
+ | (red shows lattice points within the triangle) | ||
− | ( | + | If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is <math>16</math> squares <math>(y=5.1*\pi)</math>, and the limit for the <math>x</math>-value is <math>5</math> squares. First we count the <math>1*1</math> squares. In the rightmost column, there are <math>12</math> squares with length <math>1</math> because <math>y=4*\pi</math> generates squares from <math>(4,0)</math> to <math>(4,4\pi)</math>, and continuing on we have <math>9</math>, <math>6</math>, and <math>3</math> for <math>x</math>-values for <math>1</math>, <math>2</math>, and <math>3</math> in the equation <math>y=\pi x</math>. So there are <math>12+9+6+3 = 30</math> squares with length <math>1</math> in the figure. |
− | + | For <math>2*2</math> squares, each square takes up <math>2</math> units left and <math>2</math> units up. Squares can also overlap. For <math>2*2</math> squares, the rightmost column stretches from <math>(3,0)</math> to <math>(3,3\pi)</math>, so there are <math>8</math> squares with length <math>2</math> in a <math>2</math> by <math>9</math> box. Repeating the process, the next column stretches from <math>(2,0)</math> to <math>(2,2\pi)</math>, so there are <math>5</math> squares. Continuing and adding up in the end, there are <math>8+5+2=15</math> squares with length <math>2</math> in the figure. | |
− | + | Squares with length <math>3</math> in the rightmost column start at <math>(2,0)</math> and end at <math>(2,2\pi)</math>, so there are <math>4</math> such squares in the right column. As the left row starts at <math>(1,0)</math> and ends at <math>(1,\pi)</math> there are <math>4+1=5</math> squares with length <math>3</math>. As squares with length <math>4</math> would not fit in the triangle, the answer is <math>30+15+5</math> which is <math>\boxed{\textbf{D)}\ 50}</math>. | |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=10|num-a=12}} | {{AMC12 box|year=2016|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:04, 25 July 2023
Problem
How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line , the line
and the line
Solution
Solution by e_power_pi_times_i Revised by Kinglogic and RJ5303707
(red shows lattice points within the triangle)
If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is squares
, and the limit for the
-value is
squares. First we count the
squares. In the rightmost column, there are
squares with length
because
generates squares from
to
, and continuing on we have
,
, and
for
-values for
,
, and
in the equation
. So there are
squares with length
in the figure.
For squares, each square takes up
units left and
units up. Squares can also overlap. For
squares, the rightmost column stretches from
to
, so there are
squares with length
in a
by
box. Repeating the process, the next column stretches from
to
, so there are
squares. Continuing and adding up in the end, there are
squares with length
in the figure.
Squares with length in the rightmost column start at
and end at
, so there are
such squares in the right column. As the left row starts at
and ends at
there are
squares with length
. As squares with length
would not fit in the triangle, the answer is
which is
.
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.