Difference between revisions of "2012 AMC 10B Problems/Problem 25"
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{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #22]] and [[2012 AMC 10B Problems|2012 AMC 10B #25]]}} | {{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #22]] and [[2012 AMC 10B Problems|2012 AMC 10B #25]]}} | ||
− | + | ==Problem 25== | |
A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there? | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there? | ||
Line 46: | Line 46: | ||
<math> \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 </math> | <math> \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 </math> | ||
− | ==Solution== | + | ==Solution 1== |
Line 90: | Line 90: | ||
filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);</asy> | filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);</asy> | ||
− | There is <math>1</math> way to get to any of the red arrows. From the first red arrow, there are <math>2</math> ways to get to each of the first and the second blue arrows; from the second red arrow, there are <math>3</math> ways to get to each of the first and the second blue arrows. So there are in total <math>5</math> ways to get to each of the blue arrows. | + | There is <math>1</math> way to get to any of the red arrows. From the first (top) red arrow, there are <math>2</math> ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are <math>3</math> ways to get to each of the first and the second blue arrows. So there are in total <math>5</math> ways to get to each of the blue arrows. |
From each of the first and second blue arrows, there are respectively <math>4</math> ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively <math>8</math> ways to get to each of the first and the second green arrows. Therefore there are in total <math>5 \cdot (4+4+8+8) = 120</math> ways to get to each of the green arrows. | From each of the first and second blue arrows, there are respectively <math>4</math> ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively <math>8</math> ways to get to each of the first and the second green arrows. Therefore there are in total <math>5 \cdot (4+4+8+8) = 120</math> ways to get to each of the green arrows. | ||
− | Finally, from each of the first and second green arrows, there | + | Finally, from each of the first and second green arrows, there are respectively <math>2</math> ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are <math>3</math> ways to get to the first orange arrow. Therefore there are <math>120 \cdot (2+2+3+3) = 1200</math> ways to get to each of the orange arrows, hence <math>2400</math> ways to get to the point <math>B</math>. <math>\boxed{\textbf{(E)}\ 2400}</math> |
+ | |||
+ | ==Solution 2 (using the answer choices)== | ||
+ | |||
+ | <asy> | ||
+ | size(6cm); | ||
+ | draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)); | ||
+ | draw((0.0,0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)); | ||
+ | draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)); | ||
+ | draw((4.0, 0.0)--(6.0,0.0)); | ||
+ | draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)); | ||
+ | draw((3.0,1.7320508075688772)--(4.0,-0.0)--(3.0,-1.7320508075688772), red); | ||
+ | draw((7.0,1.7320508075688772)--(6.0,-0.0)--(7.0,-1.7320508075688772), blue); | ||
+ | |||
+ | |||
+ | dot((0,0)); | ||
+ | label("$A$",(0,0),WNW); | ||
+ | filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red); | ||
+ | filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); | ||
+ | filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue); | ||
+ | filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); | ||
+ | filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue); | ||
+ | filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue); | ||
+ | filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red); | ||
+ | filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); | ||
+ | filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue); | ||
+ | </asy> | ||
+ | |||
+ | For every blue arrow, there are <math>2\cdot 2=4</math> ways to reach it without using the reverse arrow since the bug can choose any of <math>2</math> red arrows to pass through and <math>2</math> black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black arrow. Otherwise, the bug will have to travel through both red segments, which is impossible because now there is no path to take after the bug emerges from the reverse arrow. Similarly, with the blue segments, the second black arrow taken must be the one that is closest to the blue arrow that was taken. Also, it is trivial that the two black arrows taken must be different. Therefore, if the reverse arrow is taken, the blue arrow taken determines the entire path and there is <math>1</math> path for every arrow. Since the bug cannot return once it takes a blue arrow, the answer must be divisible by <math>5</math>. <math>\boxed{\textbf{(E)}\ 2400}</math> is the only answer that is. | ||
+ | |||
+ | ==Solution 3== | ||
+ | We use casework. | ||
+ | The main thing to notice is that, if the bug does not go backwards, then every vertical set of arrows can be used, as the bug could travel straight downwards and then use any arrow to continue right. | ||
+ | |||
+ | Note: The motivation is quite weird so follow my numbers as they start from the left (point A) and go right (point B). | ||
+ | |||
+ | Case 1: Bug does not go backwards. | ||
+ | |||
+ | The number of cases for this is just each vertical set of arrows multiplied to each other (if you don't understand where I'm coming from, try to understand where these numbers come from!) | ||
+ | |||
+ | And so we have <math>2*2*4*4*4*2*2 = 2^{10}</math> cases. | ||
+ | |||
+ | Case 2: The bug goes backwards once, either at the first arrow or third arrow. | ||
+ | |||
+ | Here, we have to count the fact that there is a horizontal midline that the bug could not cross, or otherwise it would be stepping on the same edge twice. | ||
+ | Back on first arrow: <math>2*1*2*4*4*2*2 = 2^{8}</math> cases. | ||
+ | Back on third arrow: <math>2*2*4*4*4*1 = 2^{8}</math> cases. | ||
+ | |||
+ | Case 3: The bug goes backwards once, at the second arrow. | ||
+ | |||
+ | Same thing as above, except since there are 4 arrows in the vertical set (plus one for the backwards arrow), then the calculations are a bit different. | ||
+ | |||
+ | We have <math>2*2*4*1*2*4*2*2 = 2^{9}</math> cases. | ||
+ | |||
+ | Notice that the first and third back arrows decrease the number of cases by a factor of <math>2^2</math> and the second back arrow decreases the number of cases by <math>2^1</math>. Hence, | ||
+ | |||
+ | 1st + 2nd = <math>2^7</math> | ||
+ | |||
+ | 2nd + 3rd = <math>2^7</math> | ||
+ | |||
+ | 1st + 3rd = <math>2^6</math> | ||
+ | |||
+ | 1st + 2nd + 3rd = <math>2^5</math> | ||
+ | |||
+ | And so the number of cases in total is <math>2^{10} + 2^9 + 2^8 + 2^8 + 2^7 + 2^7 + 2^6 + 2^5 \Rightarrow 1024 + 512 + 256 + 256 + 128 + 128 + 64 + 32 = \boxed{2400}</math> | ||
+ | |||
+ | Solution by IronicNinja | ||
+ | |||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2012amc10b/270 | ||
+ | |||
+ | ~dolphin7 | ||
== See Also == | == See Also == | ||
− | {{AMC12 box|year=2012|ab=B|num-b= | + | {{AMC10 box|year=2012|ab=B|num-b=24|after=Last Question}} |
+ | |||
+ | {{AMC12 box|year=2012|ab=B|num-b=21|num-a=23}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:06, 3 January 2021
- The following problem is from both the 2012 AMC 12B #22 and 2012 AMC 10B #25, so both problems redirect to this page.
Contents
Problem 25
A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
Solution 1
There is way to get to any of the red arrows. From the first (top) red arrow, there are ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are ways to get to each of the first and the second blue arrows. So there are in total ways to get to each of the blue arrows.
From each of the first and second blue arrows, there are respectively ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively ways to get to each of the first and the second green arrows. Therefore there are in total ways to get to each of the green arrows.
Finally, from each of the first and second green arrows, there are respectively ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are ways to get to the first orange arrow. Therefore there are ways to get to each of the orange arrows, hence ways to get to the point .
Solution 2 (using the answer choices)
For every blue arrow, there are ways to reach it without using the reverse arrow since the bug can choose any of red arrows to pass through and black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black arrow. Otherwise, the bug will have to travel through both red segments, which is impossible because now there is no path to take after the bug emerges from the reverse arrow. Similarly, with the blue segments, the second black arrow taken must be the one that is closest to the blue arrow that was taken. Also, it is trivial that the two black arrows taken must be different. Therefore, if the reverse arrow is taken, the blue arrow taken determines the entire path and there is path for every arrow. Since the bug cannot return once it takes a blue arrow, the answer must be divisible by . is the only answer that is.
Solution 3
We use casework. The main thing to notice is that, if the bug does not go backwards, then every vertical set of arrows can be used, as the bug could travel straight downwards and then use any arrow to continue right.
Note: The motivation is quite weird so follow my numbers as they start from the left (point A) and go right (point B).
Case 1: Bug does not go backwards.
The number of cases for this is just each vertical set of arrows multiplied to each other (if you don't understand where I'm coming from, try to understand where these numbers come from!)
And so we have cases.
Case 2: The bug goes backwards once, either at the first arrow or third arrow.
Here, we have to count the fact that there is a horizontal midline that the bug could not cross, or otherwise it would be stepping on the same edge twice. Back on first arrow: cases. Back on third arrow: cases.
Case 3: The bug goes backwards once, at the second arrow.
Same thing as above, except since there are 4 arrows in the vertical set (plus one for the backwards arrow), then the calculations are a bit different.
We have cases.
Notice that the first and third back arrows decrease the number of cases by a factor of and the second back arrow decreases the number of cases by . Hence,
1st + 2nd =
2nd + 3rd =
1st + 3rd =
1st + 2nd + 3rd =
And so the number of cases in total is
Solution by IronicNinja
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc10b/270
~dolphin7
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 10 Problems and Solutions |
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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.