2023 AMC 12A Problems/Problem 21
- The following problem is from both the 2023 AMC 10A #25 and 2023 AMC 12A #21, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Diagram
- 3 1 minute solution by MegaMath
- 4 Solution 1
- 5 Solution 2 (Cheese + Actual way)
- 6 Solution 3
- 7 Solution 4
- 8 Solution 5
- 9 Solution 6 (Case Work)
- 10 Solution 7 (efficient)
- 11 Solution 8 (Dodecahedron Schlegel Diagram)
- 12 Solution 9 (Symmetry)
- 13 Video Solution
- 14 Video Solution by OmegaLearn
- 15 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 16 Video Solution by epicbird08
- 17 Vide Solution by SpreadTheMathLove(Casework and Complementary)
- 18 Video Solution
- 19 Video Solution by TheBeautyofMath
- 20 See also
Problem
If and are vertices of a polyhedron, define the distance to be the minimum number of edges of the polyhedron one must traverse in order to connect and . For example, if is an edge of the polyhedron, then , but if and are edges and is not an edge, then . Let , , and be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that ?
Diagram
1 minute solution by MegaMath
https://www.youtube.com/watch?v=dxYw1wYHid4&t=12s
Solution 1
To find the total amount of vertices we first find the amount of edges, and that is . Next, to find the amount of vertices we can use Euler's characteristic, , and therefore the amount of vertices is
So there are ways to choose 3 distinct points.
Now, the furthest distance we can get from one point to another point in an icosahedron is 3. Which gives us a range of
With some case work, we get two cases:
Case 1:
Since we have only one way to choose Q, that is, the opposite point from R, we have one option for Q and any of the other points could work for S.
Then, we get (ways to choose R × ways to choose Q × ways to choose S)
Case 2:
We can visualize the icosahedron as 4 rows, first row with 1 vertex, second row with 5 vertices, third row with 5 vertices and fourth row with 1 vertex. We set R as the one vertex on the first row, and we have 12 options for R. Then, Q can be any of the 5 points on the third row and finally S can be one of the 5 points on the second row.
Therefore, we have (ways to choose R × ways to choose Q × ways to choose S)
Hence,
~lptoggled, edited by ESAOPS and trevian1
Solution 2 (Cheese + Actual way)
In total, there are ways to select the points. However, if we look at the denominators of , they are which are not divisors of . Also is impossible as cases like exist. The only answer choice left is
Note: this cheese is actually wrong because the total number of ways to select the points is actually as order matters, so all denominators are possible. Rather, you can arrive at the same conclusion by fixing R WLOG, leading to ways in total, which works for the original cheese. ~awesomeguy856
(Actual way)
Fix an arbitrary point, to select the rest points, there are ways. To make . Which means there are in total ways to make the distance the same. ~bluesoul
Solution 3
We can imagine the icosahedron as having 4 layers. 1 vertex at the top, 5 vertices below connected to the top vertex, 5 vertices below that which are 2 edges away from the top vertex, and one vertex at the bottom that is 3 edges away. WLOG because the icosahedron is symmetric around all vertices, we can say that R is the vertex at the top. So now, we just need to find the probability that S is on a layer closer to the top than Q. We can do casework on the layer S is on to get So the answer is .
-awesomeparrot
Solution 4
We can actually see that the probability that is the exact same as because and have no difference. (In other words, we can just swap Q and S, meaning that can be called the same probability-wise.) Therefore, we want to find the probability that .
WLOG, we can rotate the icosahedron so that R is the top of the icosahedron. Then we can divide this into 2 cases:
1. They are on the second layer
There are 5 ways to put one point, and 4 ways to put the other point such that . So, there are ways to put them on the second layer.
2. They are on the third layer
There are 5 ways to put one point, and 4 ways to put the other point such that . So, there are ways to put them on the third layer.
The total number of ways to choose P and S are (because there are 12 vertices), so the probability that is .
Therefore, the probability that is
~Ethanzhang1001
Solution 5
We know that there are faces. Each of those faces has borders (since each is a triangle), and each edge is used as a border twice (for each face on either side). Thus, there are edges.
By Euler's formula, which states that for all convex polyhedra, we know that there are vertices.
The answer can be counted by first counting the number of possible paths that will yield and dividing it by (or , depending on the approach). In either case, one will end up dividing by somewhere in the denominator. We can then hope that there will be no factor of in the numerator (which would cancel the in the denominator out), and answer the only option that has an in the denominator: .
~Technodoggo
Additional note by "Fruitz": Note that one can eliminate by symmetry if you swap the ineq sign.
Another note by "andliu766": A shorter way to find the number of vertices and edges is to use the fact that the MAA logo is an icosahedron. :)
Solution 6 (Case Work)
WLOG, let R be at the top-most vertex of the icosahedron. There are cases where .
Case 1: is at the bottom-most vertex
If is at the bottom-most vertex, no matter where is, . The probability that is at the bottom-most vertex is
Case 2: is at the second layer
If is at the second layer, must be at the first layer, for to be true. The probability that is at the second layer, and is at the first layer is
Solution 7 (efficient)
Since the icosahedron is symmetric polyhedron, we can rotate it so that R is on the topmost vertex. Since Q and S basically the same, we can first count the probability that .
There are 5 points such that . There is ways to choose Q and S in this case.
There are 5 points such that . There is ways to choose Q and S in this case.
There is 1 point such that . There is ways to choose Q and S in this case.
There are 11 points that are distinct from R. There is ways to choose Q and S. There is ways to choose Q and S such that . There is ways to choose Q and S such that . The probability that is therefore which corresponds to answer choice
~~afly
Solution 8 (Dodecahedron Schlegel Diagram)
Instead of thinking this problem as vertices of a icosahedron, think of it as faces of a dodecahedron. Our middle face is and we first choose and to be the faces with distance from the middle face. There are ways of doing that. Then we choose and to be the faces with distance from the middle face. There are ways of doing that. Our answer is therefore
~lopkiloinm
Solution 9 (Symmetry)
Note that the vertices of an icosahedron are in a configuration.
We will denote the probability of event happening as
We should also note that
Also, since and are basically the same.
Therefore, what we are trying to find is just which can easily be computed when we fix at the top vertex in the structure.
Since all the points are distinct (stated in problem), the conditions are only satisfied where and are at s of the structure.
since we have to choose one of the out of possible vertices and another one of out of two times.
And the answer is
~jjaamm
Video Solution
https://youtu.be/4PSrAbrjKVg?si=G24hhBrhUPio9SZD
~MathProblemSolvingSkills.com
Video Solution by OmegaLearn
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=1yLvQ0F5Z-E
Video Solution by epicbird08
~EpicBird08
Vide Solution by SpreadTheMathLove(Casework and Complementary)
https://www.youtube.com/watch?v=4FEwxwgbliQ
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by TheBeautyofMath
~IceMatrix
See also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.