Difference between revisions of "2020 AMC 12A Problems/Problem 2"
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The acronym AMC is shown in the rectangular grid below with grid lines spaced <math>1</math> unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMC<math>?</math> | The acronym AMC is shown in the rectangular grid below with grid lines spaced <math>1</math> unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMC<math>?</math> | ||
− | + | <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> | ||
+ | |||
<math>\textbf{(A) } 17 \qquad \textbf{(B) } 15 + 2\sqrt{2} \qquad \textbf{(C) } 13 + 4\sqrt{2} \qquad \textbf{(D) } 11 + 6\sqrt{2} \qquad \textbf{(E) } 21</math> | <math>\textbf{(A) } 17 \qquad \textbf{(B) } 15 + 2\sqrt{2} \qquad \textbf{(C) } 13 + 4\sqrt{2} \qquad \textbf{(D) } 11 + 6\sqrt{2} \qquad \textbf{(E) } 21</math> | ||
Revision as of 07:00, 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
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.