2023 AMC 10A Problems/Problem 18
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Quick solution)
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6 (Based on previous knowledge)
- 8 Solution 7 (Using Answer Choices)
- 9 Solution 8 (Dual)
- 10 Video Solution by Power Solve (easy to digest!)
- 11 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 12 Video Solution by OmegaLearn
- 13 Video Solution by TheBeautyofMath
- 14 Video Solution
- 15 See Also
Problem
A rhombic dodecahedron is a solid with congruent rhombus faces. At every vertex, or edges meet, depending on the vertex. How many vertices have exactly edges meet?
Solution 1
Note Euler's formula where . There are faces and the number of edges is because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are vertices on the figure. Now note that the sum of the degrees of all the points is twice the number of edges. Let the amount of vertices with edges. Now we know . Solving this system of equations gives so the answer is . ~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)
Solution 2 (Quick solution)
Let be the number of vertices with three edges, and be the number of vertices with four edges. Since there are edges on the polyhedron, we can see that . Then, . Notice that by testing the answer choices, (D) is the only one that yields an integer solution for . Thus, the answer is .
~Mathkiddie
Solution 3
With rhombi, there are total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have total edges.
Let be the number of vertices with edges (this is what the problem asks for) and be the number of vertices with edges. We have .
Euler's formula states that, for all convex polyhedra, . In our case, We know that is the total number of vertices as we are given that all vertices are connected to either or edges. Therefore,
We now have a system of two equations. There are many ways to solve for ; choosing one yields .
Even without Euler's formula, we can do a bit of answer guessing. From , we take mod on both sides.
We know that must be divisible by . We know that the factor of will not affect the divisibility by of , so we remove the . We know that is divisible by . Checking answer choices, the only one divisible by is indeed .
~Technodoggo ~zgahzlkw (small edits) ~ESAOPS (LaTeX)
Solution 4
Note that Euler's formula is . We know from the question. We also know because every face has edges and every edge is shared by faces. We can solve for the vertices based on this information.
Using the formula we can find: Let be the number of vertices with edges and be the number of vertices with edges. We know from the question and . The second equation is because the total number of points is because there are 12 rhombuses of vertices. Now, we just have to solve a system of equations. Our answer is simply just , which is ~musicalpenguin
Solution 5
Each of the twelve rhombi has two pairs of angles across from each other that must be congruent. If both pairs of angles occur at -point intersections, we have a grid of squares. If both occur at -point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a -point intersection and two at a -point intersection.
Since each -point intersection has adjacent rhombuses, we know the number of -point intersections must equal the number of -point intersections per rhombus times the number of rhombuses over . Since there are rhombuses and two -point intersections per rhombus, this works out to be:
Hence: ~hollph27 ~Minor edits by FutureSphinx
Solution 6 (Based on previous knowledge)
Note that a rhombic dodecahedron is formed when a cube is turned inside out (as seen here), thus there are 6 4-vertices (corresponding to each face of the cube) and 8 3-vertices (corresponding to each corner of the cube). Thus the answer is
Solution 7 (Using Answer Choices)
Let be the number of -edge vertices, and be the number of -edge vertices. The total number of vertices is . Now, we know that there are vertices, but we have overcounted. We have overcounted vertices times and overcounted vertices times. Therefore, we subtract and from and set it equal to our original number of vertices. From here, we reduce both sides modulo . The disappears, and the left hand side becomes . The right hand side is , meaning that must be divisible by . Looking at the answer choices, this is only possible for .
-DEVSAXENA
(Isn't this the same as the last half of Solution 2?)
Solution 8 (Dual)
Note that a rhombic dodecahedron is the dual of a cuboctahedron. A cuboctahedron has triangular faces, which correspond to vertices on a rhombic dodecahedron that have edges.
Video Solution by Power Solve (easy to digest!)
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=Z-OCnHUwnj0
Video Solution by OmegaLearn
Video Solution by TheBeautyofMath
https://www.youtube.com/watch?v=zvKijDeiYUs
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
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