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Difference between revisions of "2023 AMC 10A Problems/Problem 18"
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<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math>
 
<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math>
  
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==Solution 1==
  
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someone else do latex pls
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Note Euler's formula where <math>\text{Vertices}+\text{Faces}-\text{Edges}=2</math>.
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There are <math>12</math> faces.
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There are <math>24</math> edges, because there are 12 faces each with four edges and each edge is shared by two faces.
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Now we know that there are <math>2-12+24=14</math> vertices.
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Now note that the sum of the degrees of all the points is <math>24</math>(the number of edges). Let <math>x=</math> the number of vertices with <math>3</math> edges. Now we know <math>\frac{3x+4(14-x)}{2}=24</math>. Solving this equation gives <math>x = \boxed{\textbf{(D) }8}</math>.
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~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified) ~sonic12345 (Fixed typo)
  
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Solution 1
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==Solution 2 (Cheese)==
  
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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 D.
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Let <math>x</math> be the number of vertices with 3 edges, and <math>y</math> be the number of vertices with 4 edges. Since there are <math>\frac{4*12}{2}=24</math> edges on the polyhedron, we can see that <math>\frac{3x+4y}{2}=24</math>. Then, <math>3x+4y=48</math>. Notice that by testing the answer choices, (D) is the only one that yields an integer solution for <math>y</math>. Thus, the answer is <math>\boxed{\textbf{(D) }8}</math>.
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~aiden22gao
 
  
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Solution 2.
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~Mathkiddie
  
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With 12 rhombus, there are 12*4 sides. All the sides are shared by 2 faces. Thus we have 24 shared sides/ edges.
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==Solution 3==
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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.
 +
 
 +
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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Let a be the number of edges with 3 vertices and b be the number of edges with 4 verticies
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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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We got 3a + 4b = 48.
 
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With Euler's formula, v-3+f=2. v-24+12=2, v=14. Thus, a+b= 14.
 
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Solving the 2 equations, we got a = 8, b = 12.
 
  
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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 a = 8.
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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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Or with a keener number theory eye, we mod 4 on both side, leaving 3x mod 4 + 0 = 0.  Thus x, must be divisible by 4.
 
  
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~Technodoggo
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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~(\mod{4})</cmath>
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<cmath>3A\equiv0~(\mod{4})</cmath>
  
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==Solution 3==
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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) ~ESAOPS (LaTeX)
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==Solution 4==
 
Note that Euler's formula is <math>V+F-E=2</math>. We know <math>F=12</math> from the question. We also know <math>E = \frac{12 \cdot 4}{2} = 24</math> because every face has <math>4</math> edges and every edge is shared by <math>2</math> faces.  We can solve for the vertices based on this information.  
 
Note that Euler's formula is <math>V+F-E=2</math>. We know <math>F=12</math> from the question. We also know <math>E = \frac{12 \cdot 4}{2} = 24</math> because every face has <math>4</math> edges and every edge is shared by <math>2</math> faces.  We can solve for the vertices based on this information.  
  
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<cmath>f = 6</cmath>
 
<cmath>f = 6</cmath>
 
<cmath>t = 8</cmath>
 
<cmath>t = 8</cmath>
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Our answer is simply just <math>t</math>, which is <math>\fbox{(D) 8}</math>
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Our answer is simply just <math>t</math>, which is <math>\boxed{\textbf{(D) }8}</math>
 
~musicalpenguin
 
~musicalpenguin
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 +
==Solution 5==
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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 <math>4</math>-point intersections, we have a grid of squares. If both occur at <math>3</math>-point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a <math>3</math>-point intersection and two at a <math>4</math>-point intersection.
 +
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Since each <math>3</math>-point intersection has <math>3</math> adjacent rhombuses, we know the number of <math>3</math>-point intersections must equal the number of <math>3</math>-point intersections per rhombus times the number of rhombuses over <math>3</math>. Since there are <math>12</math> rhombuses and two <math>3</math>-point intersections per rhombus, this works out to be:
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<math>\frac{2\cdot12}{3}</math>
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Hence: <math>\boxed{\textbf{(D) }8}</math>
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~hollph27
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~Minor edits by FutureSphinx
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==Solution 6 (Based on previous knowledge)==
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Note that a rhombic dodecahedron is formed when a cube is turned inside out (as seen [https://en.wikipedia.org/wiki/Rhombic_dodecahedron#/media/File:R1-cube.gif 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 <math>\boxed{\textbf{(D) }8}</math>
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==Solution 7 (Using Answer Choices)==
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Let <math>m</math> be the number of <math>4</math>-edge vertices, and <math>n</math> be the number of <math>3</math>-edge vertices. The total number of vertices is <math>m+n</math>. Now, we know that there are <math>4 \cdot 12 = 48</math> vertices, but we have overcounted. We have overcounted <math>m</math> vertices <math>3</math> times and overcounted <math>n</math> vertices <math>2</math> times. Therefore, we subtract <math>3m</math> and <math>2n</math> from <math>48</math> and set it equal to our original number of vertices.
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<cmath>48 - 3m - 2n = m+n</cmath>
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<cmath>4m + 3n = 48</cmath>
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From here, we reduce both sides modulo <math>4</math>. The <math>4m</math> disappears, and the left hand side becomes <math>3n</math>. The right hand side is <math>0</math>, meaning that <math>3n</math> must be divisible by <math>4</math>. Looking at the answer choices, this is only possible for <math>n = \boxed{8}</math>.
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 +
-DEVSAXENA
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(Isn't this the same as the last half of Solution 2?)
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 +
==Solution 8 (Dual)==
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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.
 +
 +
==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==
 +
{{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}}
 +
{{MAA Notice}}

Revision as of 20:11, 7 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 $24$(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) ~sonic12345 (Fixed typo)

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
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All AMC 10 Problems and Solutions

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