2013 AMC 12B Problems/Problem 12

Problem 12

Cities $A$, $B$, $C$, $D$, and $E$ are connected by roads $\widetilde{AB}$, $\widetilde{AD}$, $\widetilde{AE}$, $\widetilde{BC}$, $\widetilde{BD}$, $\widetilde{CD}$, and $\widetilde{DE}$. How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities more than once.) [asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B); dot (C); dot (D); dot (E); label("$A$",A,S); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,N); label("$E$",E,W); guide squiggly(path g, real stepsize, real slope=45) {  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]

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18$


Solution

Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$, there is only one possible path to take out of cities $C$ or $E$. So the diagram is as follows: [asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B);  dot (D);  label("$A$",A,S); label("$B$",B,SE);  label("$D$",D,N);  draw(A--B..D..cycle); draw(A--D); draw(B--D);[/asy]

Now we proceed with casework. Remember that there are two ways to travel from $A$ to $D$, $D$ to $A$, $B$ to $D$ and $D$ to $B$.:

Case 1 $A \Rightarrow D$: From $D$, if the path returns to $A$, then the next path must go to $B\Rightarrow D \Rightarrow B$. There are $2 \cdot 1 \cdot 2 = 4$ possibilities of the path $ADABDB$. If the path goes to $D$ from $B$, then the path must continue with either $BDAB$ or $BADB$. There are $2 \cdot 2 \cdot 2 = 8$ possibilities. So, this case gives $4+8=12$ different possibilities.

Case 2 $A \Rightarrow B$: The path must continue with $BDADB$. There are $2 \cdot 2 = 4$ possibilities for this case.

Putting the two cases together gives $12+4 = \boxed{\textbf{(D)} \ 16}$

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
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

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