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Difference between revisions of "2020 AMC 12A Problems/Problem 2"

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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> (this can be deducted using <math>45-45-90</math>, pythag, trig, or just sense)
 
Each of the straight line segments have length <math>1</math> and each of the slanted line segments have length <math>\sqrt{2}</math> (this can be deducted using <math>45-45-90</math>, pythag, trig, or just sense)
  
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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
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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{2}}</math> ~quacker88
  
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You could have also just counted <math>4</math> slanted line segments and realized that the only answer choice involving <math>4\sqrt{2}</math> was <math>\boxed{\textbf{C) }13+4\sqrt{3}}</math>.
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You could have also just counted <math>4</math> slanted line segments and realized that the only answer choice involving <math>4\sqrt{2}</math> was <math>\boxed{\textbf{C) }13+4\sqrt{2}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 10:56, 1 February 2020

Problem

The acronym AMC is shown in the rectangular grid below with grid lines spaced $1$ 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]

$\textbf{(A) } 17 \qquad \textbf{(B) } 15 + 2\sqrt{2} \qquad \textbf{(C) } 13 + 4\sqrt{2} \qquad \textbf{(D) } 11 + 6\sqrt{2} \qquad \textbf{(E) } 21$

Solution

Each of the straight line segments have length $1$ and each of the slanted line segments have length $\sqrt{2}$ (this can be deducted using $45-45-90$, pythag, trig, or just sense)

There area a total of $13$ straight lines segments and $4$ slanted line segments. The sum is $\boxed{\textbf{C) }13+4\sqrt{2}}$ ~quacker88

You could have also just counted $4$ slanted line segments and realized that the only answer choice involving $4\sqrt{2}$ was $\boxed{\textbf{C) }13+4\sqrt{2}}$.

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. AMC logo.png

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