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

Revision as of 02:46, 31 May 2024

1 Problem
2 Solution 1
3 Solution 2 (Cheese)
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 $12$ 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 Power Solve (easy to digest!)

https://youtu.be/5OuzPFvJPEY

Video Solution by CosineMethod [🔥Fast and Easy🔥]

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)

@@ Line 1: / Line 1: @@
-Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
+==Problem==
+A rhombic dodecahedron is a solid with <math>12</math> congruent rhombus faces. At every vertex, <math>3</math> or <math>4</math> edges meet, depending on the vertex. How many vertices have exactly <math>3</math> edges meet?
--Jonathan Yu
+<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math>
+==Solution 1==
+Note Euler's formula where <math>\text{Vertices}+\text{Faces}-\text{Edges}=2</math>.
+There are <math>12</math> faces.
+There are <math>24</math> edges, because there are 12 faces each with four edges and each edge is shared by two faces.
+Now we know that there are <math>2-12+24=14</math> vertices.
+Now note that the sum of the degrees of all the points is <math>2</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>.
+~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)
+==Solution 2 (Cheese)==
+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>.
+~Mathkiddie
+==Solution 3==
+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>.
+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>
+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>.
+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.
+<cmath>3A+4B\equiv48~(\mod{4})</cmath>
+<cmath>3A\equiv0~(\mod{4})</cmath>
+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)
+==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.
+Using the formula we can find:
+<cmath>V + 12 - 24 = 2</cmath>
+<cmath>V = 14</cmath>
+Let <math>t</math> be the number of vertices with <math>3</math> edges and <math>f</math> be the number of vertices with <math>4</math> edges. We know <math>t+f = 14</math> from the question and <math>3t + 4f = 48</math>. The second equation is because the total number of points is <math>48</math> because there are 12 rhombuses of <math>4</math> vertices.
+Now, we just have to solve a system of equations.
+<cmath>3t + 4f = 48</cmath>
+<cmath>3t + 3f = 42</cmath>
+<cmath>f = 6</cmath>
+<cmath>t = 8</cmath>
+Our answer is simply just <math>t</math>, which is <math>\boxed{\textbf{(D) }8}</math>
+~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 <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.
+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:
+<math>\frac{2\cdot12}{3}</math>
+Hence: <math>\boxed{\textbf{(D) }8}</math>
+~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 [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>
+==Solution 7 (Using Answer Choices)==
+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.
+<cmath>48 - 3m - 2n = m+n</cmath>
+<cmath>4m + 3n = 48</cmath>
+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>.
+-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 <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 Power Solve (easy to digest!)==
+https://youtu.be/5OuzPFvJPEY
+== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
+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==
+{{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}}
+{{MAA Notice}}

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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