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, where x is the number of tiles on the side length of the square. This is because the number of tiles on the square's diagonal is equal to the number of tiles on the square's side length.Therefore, our answer is <math>19^2=\boxed{\textbf{(C)}\ 361}</math>. |
| − | ==Video Solution== | + | ~AllezW |
| − | + | ||
| + | ~Edits and Add-On's by [[User: Wrenmath|WrenMath]] | ||
| + | |||
| + | ==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 = \boxed{\textbf{(C)}\ 361}</math>. | ||
| + | |||
| + | ~Ligonmathkid2 | ||
| + | |||
| + | ~Minor Edits by [[User: Wrenmath|WrenMath]] | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING!!!)== | ||
| + | https://youtu.be/_XrCl3p--28 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution by RMM Club== | ||
| + | https://youtu.be/QCWOZwYVJMg | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
https://youtu.be/8XtEOkP-AS0 | https://youtu.be/8XtEOkP-AS0 | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
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| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 19:46, 7 June 2025
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, where x is the number of tiles on the side length of the square. This is because the number of tiles on the square's diagonal is equal to the number of tiles on the square's side length.Therefore, our answer is 
.
~AllezW
~Edits and Add-On's by WrenMath
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
~Minor Edits by WrenMath
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by RMM Club
Video Solution by WhyMath
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
