Difference between revisions of "2011 AMC 12A Problems/Problem 2"
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There are <math>5</math> coins placed flat on a table according to the figure. What is the order of the coins from top to bottom? | There are <math>5</math> coins placed flat on a table according to the figure. What is the order of the coins from top to bottom? | ||
| − | <math> | + | <asy> |
| − | \textbf{(A)}\ (C, A, E, D, B) \qquad | + | size(100); defaultpen(linewidth(.8pt)+fontsize(8pt)); |
| − | \textbf{(B)}\ (C, A, D, E, B) \qquad | + | draw(arc((0,1), 1.2, 25, 214)); |
| − | \textbf{(C)}\ (C, D, E, A, B) \qquad | + | draw(arc((.951,.309), 1.2, 0, 360)); |
| − | \textbf{(D)}\ (C, E, A, D, B) \qquad | + | draw(arc((.588,-.809), 1.2, 132, 370)); |
| − | \textbf{(E)}\ (C, E, D, A, B) </math> | + | draw(arc((-.588,-.809), 1.2, 75, 300)); |
| + | draw(arc((-.951,.309), 1.2, 96, 228)); | ||
| + | label("$A$",(0,1),NW); label("$B$",(-1.1,.309),NW); label("$C$",(.951,.309),E); label("$D$",(-.588,-.809),W); label("$E$",(.588,-.809),S); | ||
| + | </asy> | ||
| + | |||
| + | <math>\textbf{(A)}\ (C, A, E, D, B) \qquad \textbf{(B)}\ (C, A, D, E, B) \qquad \textbf{(C)}\ (C, D, E, A, B) \qquad \textbf{(D)}\ (C, E, A, D, B) \qquad \textbf{(E)}\ (C, E, D, A, B) </math> | ||
== Solution == | == Solution == | ||
| + | By careful inspection and common sense, the answer is <math>\textbf{(E)}</math>. | ||
| + | |||
| + | == Note == | ||
| + | Note that this image is quite similar to a topological map. This means that the coin that appears closest to us is at the top and the coin that appears farthest from us is at the bottom. | ||
| + | |||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=1|num-a=3|ab=A}} | {{AMC12 box|year=2011|num-b=1|num-a=3|ab=A}} | ||
| + | |||
| + | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:00, 25 September 2025
Contents
Problem
There are 
coins placed flat on a table according to the figure. What is the order of the coins from top to bottom?
![[asy] size(100); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw(arc((0,1), 1.2, 25, 214)); draw(arc((.951,.309), 1.2, 0, 360)); draw(arc((.588,-.809), 1.2, 132, 370)); draw(arc((-.588,-.809), 1.2, 75, 300)); draw(arc((-.951,.309), 1.2, 96, 228)); label("$A$",(0,1),NW); label("$B$",(-1.1,.309),NW); label("$C$",(.951,.309),E); label("$D$",(-.588,-.809),W); label("$E$",(.588,-.809),S); [/asy]](http://latex.artofproblemsolving.com/b/f/5/bf5741398e1d7e23c76474c5b6c6c4148784f554.png)
Solution
By careful inspection and common sense, the answer is 
.
Note
Note that this image is quite similar to a topological map. This means that the coin that appears closest to us is at the top and the coin that appears farthest from us is at the bottom.
See also
| 2011 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, as part of the American Mathematics Competitions. 
