2019 AMC 12B Problems/Problem 10

Revision as of 17:13, 14 February 2019 by Arpitr20 (talk | contribs) (Solution)

Problem

The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$, without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.)

!! Someone with good Latex or Asymptote skills please help !!

How many different routes can Paula take?

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution

Note the following route, which isn't that hard to discover:

!! Someone with good Latex/Asymptote skills please help !!

Look at the two square "loop"s. Each one can be oriented in one of two directions (lower left: either go down or left first; upper right: either go right or up first). Therefore, the answer is 1 route * 2 * 2 = $\boxed{\textbf{(E) } 4}$. Note that no choice is larger than it.

Solution 2

Note that of the 12 cities, 6 of them (2 on the top and bottom, 1 on the sides) have 3 edges connecting to them. Therefore, at least 1 edge connecting to them cannot be used. Additionally, the same goes for the start and end point as we don't want to return to them. so we have 8 points that we know have 1 unused edge, and we have a total of 4 unused edges to work with (17-13), so we easily find there is only one configuration that satisfies this:


X's represent unused edges, by necessity, lines are filled in for the paths. o _ _ _ o X o _ _ _ o X | | | o _ _ _ 0 _ _ _ 0 _ _ _ o | | | X o _ _ _ o X o _ _ _ o


Now, we find that at each of the 2 cities marked with a 0, we have 2 possibilities to follow the path, we can either continue straight and cross back over it later, or make a left turn and turn right when we approach the junction again/ This gives us $2*2=\boxed4$

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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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