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Difference between revisions of "2023 AMC 10A Problems/Problem 18"

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==Solution 8 (Dual)==
 
==Solution 8 (Dual)==
 
Note that a rhombic dodecahedron is the dual of a cuboctahedron. A cuboctahedron has <math>8</math> triangular faces, which correspond to <math>\boxed{\textbf{(D) }8}</math> vertices on a rhombic dodecahedron that have <math>3</math> edges.  
 
Note that a rhombic dodecahedron is the dual of a cuboctahedron. A cuboctahedron has <math>8</math> triangular faces, which correspond to <math>\boxed{\textbf{(D) }8}</math> vertices on a rhombic dodecahedron that have <math>3</math> edges.  
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==Video Solution by Math-X (First fully understand the problem!!!)==
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https://youtu.be/GP-DYudh5qU?si=fFif-OiVZnkdTTv0&t=5105
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~Math-X
  
 
==Video Solution ==
 
==Video Solution ==

Revision as of 06:20, 6 September 2024

Problem

A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?

$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution 1

Note Euler's formula where $\text{Vertices}+\text{Faces}-\text{Edges}=2$. There are $12$ faces. There are $24$ edges, because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are $2-12+24=14$ vertices. Now note that the sum of the degrees of all the points is $2$(the number of edges). Let $x=$ the number of vertices with $3$ edges. Now we know $\frac{3x+4(14-x)}{2}=24$. Solving this equation gives $x = \boxed{\textbf{(D) }8}$. ~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)

Solution 2 (Cheese)

Let $x$ be the number of vertices with 3 edges, and $y$ be the number of vertices with 4 edges. Since there are $\frac{4*12}{2}=24$ edges on the polyhedron, we can see that $\frac{3x+4y}{2}=24$. Then, $3x+4y=48$. Notice that by testing the answer choices, (D) is the only one that yields an integer solution for $y$. Thus, the answer is $\boxed{\textbf{(D) }8}$.

~Mathkiddie

Solution 3

With $12$ rhombi, there are $4\cdot12=48$ total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have $\dfrac{48}2=24$ total edges.

Let $A$ be the number of vertices with $3$ edges (this is what the problem asks for) and $B$ be the number of vertices with $4$ edges. We have $3A + 4B = 48$.

Euler's formula states that, for all convex polyhedra, $V-E+F=2$. In our case, $V-24+12=2\implies V=14.$ We know that $A+B$ is the total number of vertices as we are given that all vertices are connected to either $3$ or $4$ edges. Therefore, $A+B=14.$

We now have a system of two equations. There are many ways to solve for $A$; choosing one yields $A=\boxed{\textbf{(D) }8}$.

Even without Euler's formula, we can do a bit of answer guessing. From $3A+4B=48$, we take mod $4$ on both sides.

\[3A+4B\equiv48~(\mod{4})\] \[3A\equiv0~(\mod{4})\]

We know that $3A$ must be divisible by $4$. We know that the factor of $3$ will not affect the divisibility by $4$ of $3A$, so we remove the $3$. We know that $A$ is divisible by $4$. Checking answer choices, the only one divisible by $4$ is indeed $A=\boxed{\textbf{(D) }8}$.

~Technodoggo ~zgahzlkw (small edits) ~ESAOPS (LaTeX)

Solution 4

Note that Euler's formula is $V+F-E=2$. We know $F=12$ from the question. We also know $E = \frac{12 \cdot 4}{2} = 24$ because every face has $4$ edges and every edge is shared by $2$ faces. We can solve for the vertices based on this information.

Using the formula we can find: \[V + 12 - 24 = 2\] \[V = 14\] Let $t$ be the number of vertices with $3$ edges and $f$ be the number of vertices with $4$ edges. We know $t+f = 14$ from the question and $3t + 4f = 48$. The second equation is because the total number of points is $48$ because there are 12 rhombuses of $4$ vertices. Now, we just have to solve a system of equations. \[3t + 4f = 48\] \[3t + 3f = 42\] \[f = 6\] \[t = 8\] Our answer is simply just $t$, which is $\boxed{\textbf{(D) }8}$ ~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 $4$-point intersections, we have a grid of squares. If both occur at $3$-point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a $3$-point intersection and two at a $4$-point intersection.

Since each $3$-point intersection has $3$ adjacent rhombuses, we know the number of $3$-point intersections must equal the number of $3$-point intersections per rhombus times the number of rhombuses over $3$. Since there are $12$ rhombuses and two $3$-point intersections per rhombus, this works out to be:

$\frac{2\cdot12}{3}$

Hence: $\boxed{\textbf{(D) }8}$ ~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 $\boxed{\textbf{(D) }8}$

Solution 7 (Using Answer Choices)

Let $m$ be the number of $4$-edge vertices, and $n$ be the number of $3$-edge vertices. The total number of vertices is $m+n$. Now, we know that there are $4 \cdot 12 = 48$ vertices, but we have overcounted. We have overcounted $m$ vertices $3$ times and overcounted $n$ vertices $2$ times. Therefore, we subtract $3m$ and $2n$ from $48$ and set it equal to our original number of vertices. \[48 - 3m - 2n = m+n\] \[4m + 3n = 48\] From here, we reduce both sides modulo $4$. The $4m$ disappears, and the left hand side becomes $3n$. The right hand side is $0$, meaning that $3n$ must be divisible by $4$. Looking at the answer choices, this is only possible for $n = \boxed{8}$.

-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 $8$ triangular faces, which correspond to $\boxed{\textbf{(D) }8}$ vertices on a rhombic dodecahedron that have $3$ edges.

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/GP-DYudh5qU?si=fFif-OiVZnkdTTv0&t=5105

~Math-X

Video Solution

https://youtu.be/5OuzPFvJPEY

Video Solution

https://www.youtube.com/watch?v=Z-OCnHUwnj0

Video Solution by OmegaLearn

https://youtu.be/0AG5XmWY-D8

Video Solution by TheBeautyofMath

https://www.youtube.com/watch?v=zvKijDeiYUs

Video Solution

https://youtu.be/0ssjr8KjOzk

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also Cheese

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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