Difference between revisions of "2023 AMC 12A Problems/Problem 21"
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+ | ==Video Solution== | ||
+ | https://youtu.be/4PSrAbrjKVg?si=G24hhBrhUPio9SZD | ||
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+ | ~MathProblemSolvingSkills.com | ||
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==Video Solution by OmegaLearn== | ==Video Solution by OmegaLearn== |
Revision as of 20:10, 25 November 2023
- 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 Solution 1
- 3 Solution 2 (Cheese + Actual way)
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6 (Case Work)
- 8 Solution 7 (efficient)
- 9 Solution 8 (Dodecahedron Schlegel Diagram)
- 10 Video Solution
- 11 Video Solution by OmegaLearn
- 12 1 minute solution by MegaMath
- 13 Video Solution by epicbird08
- 14 Vide Solution by SpreadTheMathLove(Casework and Complementary)
- 15 Video Solution
- 16 Video Solution by TheBeautyofMath
- 17 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
?
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
Video Solution
https://youtu.be/4PSrAbrjKVg?si=G24hhBrhUPio9SZD
~MathProblemSolvingSkills.com
Video Solution by OmegaLearn
1 minute solution by MegaMath
https://www.youtube.com/watch?v=dxYw1wYHid4&t=12s
~megahertz13
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.