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

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==Solution 2==
 
==Solution 2==
  
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With 12 rhombi, there are <math>48</math> sides. All the sides are shared by 2 faces. Thus we have <math>24</math> shared sides/edges.
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With <math>12</math> rhombi, there are <math>4\cdot12=48</math> total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have <math>\dfrac{48}2=24</math> total edges.  
  
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Let <math>A</math> be the number of edges with 3 vertices and <math>B</math> be the number of edges with 4 vertices.
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Let <math>A</math> be the number of vertices with <math>3</math> edges (this is what the problem asks for) and <math>B</math> be the number of vertices with <math>4</math> edges. We have <math>3A + 4B = 48</math>.  
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We get <math>3A + 4B = 48</math>.
 
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With Euler's formula, <math>V-3+F=2</math>.  <math>V-24+12=2</math>, so <math>V = 14</math>. Thus, <math>a+b= 14</math>.
 
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Solving the 2 equations, we get <math>a = 8</math> and <math>b = 6</math>.
 
  
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Even without Euler's formula, we observe that a must be even integers, so trying even integer choices and we also get <math>a = \boxed{\textbf{(D) }8}</math>.
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Euler's formula states that, for all convex polyhedra, <math>v-e+f=2</math>. In our case, <math>v-24+12=2\implies v=14.</math> We know that <math>A+B</math> is the total number of vertices as we are given that all vertices are connected to either <math>3</math> or <math>4</math> edges. Therefore, <math>A+B=14.</math>
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Or with a keener number theory eye, we mod 4 on both side, leaving <math>3x</math> mod <math>4 + 0 = 0</math>. Thus, <math>x</math> must be divisible by 4.
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 +
We now have a system of two equations. There are many ways to solve for <math>A</math>; choosing one yields <math>A=\boxed}\textbf{(D) }8</math>.  
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Even without Euler's formula, we can do a bit of answer guessing. From <math>3A+4B=48</math>, we take mod <math>4</math> on both sides.
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<cmath>3A+4B\equiv48~(\mod4)</cmath>
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<cmath>3A\equiv0~(\mod4)</cmath>
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 +
We know that <math>3A</math> must be divisible by <math>4</math>. We know that the factor of <math>3</math> will not affect the divisibility by <math>4</math> of <math>3A</math>, so we remove the <math>3</math>. We know that <math>A</math> is divisible by <math>4</math>. Checking answer choices, the only one divisible by <math>4</math> is indeed <math>A=\boxed}\textbf{(D) }8</math>.  
  
 
~Technodoggo ~zgahzlkw (small edits)
 
~Technodoggo ~zgahzlkw (small edits)

Revision as of 22:08, 9 November 2023

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 $V+F-E=2$. There are $12$ faces and the number of edges is $24$ because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are $14$ vertices on the figure. Let $A$ be the number of vertices with degree 3 and $B$ be the number of vertices with degree 4. $A+B=14$ is our first equation. Now note that the sum of the degrees of all the points is twice the number of edges. Now we know $3A+4B=48$. Solving this system of equations gives $B = 6$ and $A = 8$ so the answer is $\boxed{\textbf{(D) }8}$. ~aiden22gao ~zgahzlkw (LaTeX)

Solution 2

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 $A=\boxed}\textbf{(D) }8$ (Error compiling LaTeX. Unknown error_msg).

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

\[3A+4B\equiv48~(\mod4)\] \[3A\equiv0~(\mod4)\]

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 $A=\boxed}\textbf{(D) }8$ (Error compiling LaTeX. Unknown error_msg).

~Technodoggo ~zgahzlkw (small edits)

Solution 3

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 4

Each of the twelve rhombuses has two pairs of angles across from each other that must be the same. 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:

$\frac{2*12}{3}$

Hence: $\boxed{\textbf{(D) }8}$ ~hollph27

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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