Difference between revisions of "2017 AMC 8 Problems/Problem 11"
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==Solution== | ==Solution== | ||
Since the number of tiles lying on both diagonals is <math>37</math>, counting one tile twice, there are <math>37=2x-1\implies x=19</math> tiles on each side. Hence, our answer is <math>19^2=361=\boxed{\textbf{(C)}\ 361}</math>. | Since the number of tiles lying on both diagonals is <math>37</math>, counting one tile twice, there are <math>37=2x-1\implies x=19</math> tiles on each side. Hence, our answer is <math>19^2=361=\boxed{\textbf{(C)}\ 361}</math>. | ||
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| + | ==Solution 2== | ||
| + | Visualize it as 4 separate diagonals connecting to one square in the middle. Each square on the diagonal corresponds to one square of horizontal/vertical distance (because it's a square). So we figure out the length of each separate diagonal, multiply by two, and then add 1 (realize that we can just join two of the separate diagonals on opposite sides together to save some time in calculations). Therefore, the edge length is <cmath>\frac{37-1}{4} \cdot 2 + 1 = 19</cmath> | ||
| + | Thus, our solution is <math>19^2 = 361 = \boxed{\textbf{(C)}\ 361}</math>. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 10:49, 17 February 2023
Problem 11
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Solution
Since the number of tiles lying on both diagonals is 
, counting one tile twice, there are 
tiles on each side. Hence, our answer is 
.
Solution 2
Visualize it as 4 separate diagonals connecting to one square in the middle. Each square on the diagonal corresponds to one square of horizontal/vertical distance (because it's a square). So we figure out the length of each separate diagonal, multiply by two, and then add 1 (realize that we can just join two of the separate diagonals on opposite sides together to save some time in calculations). Therefore, the edge length is ![\[\frac{37-1}{4} \cdot 2 + 1 = 19\]](http://latex.artofproblemsolving.com/e/3/a/e3a67692ffe3dc442c8fc30d6d616f51fa32888a.png)
Thus, our solution is 
.
Video Solution
Associated video: https://youtu.be/QCWOZwYVJMg
~savannahsolver
See Also:
| 2017 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
