Difference between revisions of "2013 AMC 12B Problems/Problem 12"
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− | ==Problem== | + | == Problem == |
+ | Cities <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are connected by roads <math>\widetilde{AB}</math>, <math>\widetilde{AD}</math>, <math>\widetilde{AE}</math>, <math>\widetilde{BC}</math>, <math>\widetilde{BD}</math>, <math>\widetilde{CD}</math>, and <math>\widetilde{DE}</math>. How many different routes are there from <math>A</math> to <math>B</math> that use each road exactly once? (Such a route will necessarily visit some cities more than once.) | ||
− | |||
<asy> | <asy> | ||
unitsize(10mm); | unitsize(10mm); | ||
Line 17: | Line 17: | ||
label("$D$",D,N); | label("$D$",D,N); | ||
label("$E$",E,W); | label("$E$",E,W); | ||
− | draw(A--B | + | guide squiggly(path g, real stepsize, real slope=45) |
− | draw( | + | { |
− | draw( | + | real len = arclength(g); |
+ | real step = len / round(len / stepsize); | ||
+ | guide squig; | ||
+ | for (real u = 0; u < len; u += step){ | ||
+ | real a = arctime(g, u); | ||
+ | real b = arctime(g, u + step / 2); | ||
+ | pair p = point(g, a); | ||
+ | pair q = point(g, b); | ||
+ | pair np = unit( rotate(slope) * dir(g,a)); | ||
+ | pair nq = unit( rotate(0 - slope) * dir(g,b)); | ||
+ | squig = squig .. p{np} .. q{nq}; | ||
+ | } | ||
+ | squig = squig .. point(g, length(g)){unit(rotate(slope)*dir(g,length(g)))}; | ||
+ | return squig; | ||
+ | } | ||
+ | pen pp = defaultpen + 2.718; | ||
+ | draw(squiggly(A--B, 4.04, 30), pp); | ||
+ | draw(squiggly(A--D, 7.777, 20), pp); | ||
+ | draw(squiggly(A--E, 5.050, 15), pp); | ||
+ | draw(squiggly(B--C, 5.050, 15), pp); | ||
+ | draw(squiggly(B--D, 4.04, 20), pp); | ||
+ | draw(squiggly(C--D, 2.718, 20), pp); | ||
+ | draw(squiggly(D--E, 2.718, -60), pp); | ||
+ | </asy> | ||
<math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18</math> | <math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18</math> | ||
− | ==Solution== | + | == Solution == |
− | |||
Note that cities <math>C</math> and <math>E</math> can be removed when counting paths because if a path goes in to <math>C</math> or <math>E</math>, there is only one possible path to take out of cities <math>C</math> or <math>E</math>. | Note that cities <math>C</math> and <math>E</math> can be removed when counting paths because if a path goes in to <math>C</math> or <math>E</math>, there is only one possible path to take out of cities <math>C</math> or <math>E</math>. | ||
So the diagram is as follows: | So the diagram is as follows: | ||
+ | |||
<asy> | <asy> | ||
unitsize(10mm); | unitsize(10mm); | ||
Line 44: | Line 67: | ||
draw(A--B..D..cycle); | draw(A--B..D..cycle); | ||
draw(A--D); | draw(A--D); | ||
− | draw(B--D);</asy> | + | draw(B--D); |
+ | </asy> | ||
Now we proceed with casework. Remember that there are two ways to travel from <math>A</math> to <math>D</math>, <math>D</math> to <math>A</math>, <math>B</math> to <math>D</math> and <math>D</math> to <math>B</math>.: | Now we proceed with casework. Remember that there are two ways to travel from <math>A</math> to <math>D</math>, <math>D</math> to <math>A</math>, <math>B</math> to <math>D</math> and <math>D</math> to <math>B</math>.: | ||
− | Case 1 <math>A \Rightarrow D</math>: From <math>D</math>, if the path returns to <math>A</math>, then the next path must go to <math>B\Rightarrow D \Rightarrow B</math>. There are <math>2 \cdot 1 \cdot 2 = 4</math> possibilities of the path <math>ADABDB</math>. If the path goes to <math> | + | Case 1 <math>A \Rightarrow D</math>: From <math>D</math>, if the path returns to <math>A</math>, then the next path must go to <math>B\Rightarrow D \Rightarrow B</math>. There are <math>2 \cdot 1 \cdot 2 = 4</math> possibilities of the path <math>ADABDB</math>. If the path goes to <math>D</math> from <math>B</math>, then the path must continue with either <math>BDAB</math> or <math>BADB</math>. There are <math>2 \cdot 2 \cdot 2 = 8</math> possibilities. So, this case gives <math>4+8=12</math> different possibilities. |
Case 2 <math>A \Rightarrow B</math>: The path must continue with <math>BDADB</math>. There are <math>2 \cdot 2 = 4</math> possibilities for this case. | Case 2 <math>A \Rightarrow B</math>: The path must continue with <math>BDADB</math>. There are <math>2 \cdot 2 = 4</math> possibilities for this case. | ||
− | Putting the two cases together gives <math>12+4 = \boxed{\textbf{(D)}16}</math> | + | Putting the two cases together gives <math>12+4 = \boxed{\textbf{(D)} \ 16}</math> |
− | == See | + | == See Also == |
{{AMC12 box|year=2013|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2013|ab=B|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Latest revision as of 22:11, 3 October 2022
Problem
Cities ,
,
,
, and
are connected by roads
,
,
,
,
,
, and
. How many different routes are there from
to
that use each road exactly once? (Such a route will necessarily visit some cities more than once.)
Solution
Note that cities and
can be removed when counting paths because if a path goes in to
or
, there is only one possible path to take out of cities
or
.
So the diagram is as follows:
Now we proceed with casework. Remember that there are two ways to travel from to
,
to
,
to
and
to
.:
Case 1 : From
, if the path returns to
, then the next path must go to
. There are
possibilities of the path
. If the path goes to
from
, then the path must continue with either
or
. There are
possibilities. So, this case gives
different possibilities.
Case 2 : The path must continue with
. There are
possibilities for this case.
Putting the two cases together gives
See Also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.