Difference between revisions of "2020 AMC 12A Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
Either count the straight or diagonals and deduce from the answers that the only answer possible is <math>\boxed{\textbf{C) }13+4\sqrt{2}}</math>. | Either count the straight or diagonals and deduce from the answers that the only answer possible is <math>\boxed{\textbf{C) }13+4\sqrt{2}}</math>. | ||
| + | |||
| + | ==Solution 3== | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 20:19, 24 March 2021
Problem
The acronym AMC is shown in the rectangular grid below with grid lines spaced 
unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMC
![[asy] import olympiad; unitsize(25); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 9; ++j) { pair A = (j,i); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 9; ++j) { if (j != 8) { draw((j,i)--(j+1,i), dashed); } if (i != 2) { draw((j,i)--(j,i+1), dashed); } } } draw((0,0)--(2,2),linewidth(2)); draw((2,0)--(2,2),linewidth(2)); draw((1,1)--(2,1),linewidth(2)); draw((3,0)--(3,2),linewidth(2)); draw((5,0)--(5,2),linewidth(2)); draw((4,1)--(3,2),linewidth(2)); draw((4,1)--(5,2),linewidth(2)); draw((6,0)--(8,0),linewidth(2)); draw((6,2)--(8,2),linewidth(2)); draw((6,0)--(6,2),linewidth(2)); [/asy]](http://latex.artofproblemsolving.com/6/6/4/6649eb6b62018218d4086bde7597f2da9b4b10e3.png)
Solution 1
Each of the straight line segments have length and each of the slanted line segments have length 
(this can be deducted using 
, pythag, trig, or just sense)
There area a total of 
straight lines segments and 
slanted line segments. The sum is 
~quacker88
Solution 2
Either count the straight or diagonals and deduce from the answers that the only answer possible is .
Solution 3
Video Solution
~IceMatrix
See Also
| 2020 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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. 
