2020 AMC 12A Problems/Problem 2: Difference between revisions
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Each of the straight line segments have length <math>1</math> and each of the slanted line segments have length <math>\sqrt{2}</math>. | Each of the straight line segments have length <math>1</math> and each of the slanted line segments have length <math>\sqrt{2}</math>. | ||
There area a total of <math>13</math> straight lines segments and <math>4</math> slanted line segments. The sum is <math>\boxed{\textbf{C) }13+4\sqrt{3}</math> ~quacker88 | There area a total of <math>13</math> straight lines segments and <math>4</math> slanted line segments. The sum is <math>\boxed{\textbf{C) }13+4\sqrt{3}}</math> ~quacker88 | ||
==See Also== | ==See Also== | ||
Revision as of 10:55, 1 February 2020
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
Each of the straight line segments have length and each of the slanted line segments have length 
.
There area a total of 
straight lines segments and 
slanted line segments. The sum is 
~quacker88
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
