Difference between revisions of "2024 AMC 8 Problems/Problem 14"
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~SpeedCuber7 | ~SpeedCuber7 | ||
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*AM can be crossed off because AXM is shorter | *AM can be crossed off because AXM is shorter |
Revision as of 13:09, 3 April 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Advanced)
- 4 Solution 3 (Variant of Solution 1)
- 5 Solution 4 (On paper)
- 6 Video Solution 1 (easy to digest) by Power Solve
- 7 Video Solution 2 by SpreadTheMathLove
- 8 Video Solution 3 by NiuniuMaths (Easy to understand!)
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 Video Solution by Math-X (First fully understand the problem!!!)
- 11 Video Solution by Interstigation
- 12 See Also
Problem
The one-way routes connecting towns and are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
Solution 1
We can simply see that path will give us the smallest value. Adding, . This is nice as it’s also the smallest value, solidifying our answer.
You can also simply brute-force it or sort of think ahead - for example, getting from A to M can be done ways; () or , so you should take the shorter route (). Another example is M to C, two ways - one is and the other is . Take the shorter route. After this, you need to consider a few more times - consider if () is greater than ), which it is not, and consider if ) is greater than () or () which it is not. TL;DR: . [Note: This is probably just the thinking behind the solution.] {Double-note: As MaxyMoosy said, since this answer is the smallest one, it has to be the right answer.}
~MaxyMoosy ~HACKER2022
Solution 2 (Advanced)
We can execute Dijkstra's algorithm by hand to find the shortest path from to every other town, including . This effectively proves that, assuming we execute the algorithm correctly, that we will have found the shortest distance. The distance estimates for each step of the algorithm (from to each node) are shown below:
The steps are as follows: starting with the initial node , set and for all where indicates the distance from to . Consider the outgoing edges and and update the distance estimates and , completing the first row of the table.
The node is the unvisited node with the lowest distance estimate, so we will consider and its outgoing edges and . The distance estimate equals , and the distance estimate updates to , because . This completes the second row of the table. Repeating this process for each unvisited node (in order of its distance estimate) yields the correct distance once the algorithm is complete.
~scrabbler94
Solution 3 (Variant of Solution 1)
From , we want to find the shortest route to , so we want to try to find the shortest path through each node (not necessarily all of them). We should follow the arrows, since all of them lead to . From , there are paths we can take, to , or to . We travel to , since . From , we go to , we go to , we go to , and finally go to . Adding up gives .
~SpeedCuber7
Solution 4 (On paper)
We can cross off a few routes:
- AM can be crossed off because AXM is shorter
- XY can be crossed off because XMY is shorter
- YZ can be crossed off because YCZ is shorter
- MC can be crossed off because MYC is shorter
- MZ can be crossed off because MCZ is shorter
Finally, we are left with a single path AXMYCZ from A to Z which adding it up gives
Video Solution 1 (easy to digest) by Power Solve
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=RRTxlduaDs8
Video Solution 3 by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=bAxLRYT6SCw
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=y_YVZFC4DQdB2qhM&t=3280
~Math-X
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=1405
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.