2016 AMC 12B Problems/Problem 11
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
![[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]](http://latex.artofproblemsolving.com/a/b/c/abc2430047d924cb59d8f9799986b35cea9a979a.png)
(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. 
