Difference between revisions of "2017 AMC 8 Problems/Problem 11"
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<math>\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369</math> | <math>\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | 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. | + | 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. Therefore, our answer is <math>19^2=361=\boxed{\textbf{(C)}\ 361}</math>. |
| − | ==See Also== | + | ~AllezW |
| + | |||
| + | ==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>. | ||
| + | |||
| + | ~Ligonmathkid2 | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING!!!)== | ||
| + | https://youtu.be/_XrCl3p--28 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | Associated video: https://youtu.be/QCWOZwYVJMg | ||
| + | |||
| + | https://youtu.be/8XtEOkP-AS0 | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See Also:== | ||
{{AMC8 box|year=2017|num-b=10|num-a=12}} | {{AMC8 box|year=2017|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:01, 27 January 2024
Contents
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 1
Since the number of tiles lying on both diagonals is 
, counting one tile twice, there are 
tiles on each side. Therefore, our answer is 
.
~AllezW
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 
.
~Ligonmathkid2
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
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. 
