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Difference between revisions of "2023 AMC 10A Problems/Problem 18"
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-Note: this page was griefed by Alac 16 and I am trying my best to restore it. Formatting help would be much appreciated-Toucan2009hz
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Contents
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1 Problem
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2 Solution 1
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3 Solution 2 (Cheese)
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4 Solution 3
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5 Solution 4
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6 Solution 5
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7 Solution 6 (Based on previous knowledge)
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8 Solution 7 (Dual)
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9 Video Solution by Math-X (First fully understand the problem!!!)
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10 Video Solution
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11 Video Solution
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12 Video Solution by OmegaLearn
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13 Video Solution by TheBeautyofMath
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14 Video Solution
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15 See Also Cheese
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Problem
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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?
  
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Website: https://www.google.com/
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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>
  
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Gmail: alaccary446@gmail.com
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Solution 1
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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>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>. ~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified) ~sonic12345 (Fixed typo)
  
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Hashtag: #Google trends #google news #google map #google drive
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Solution 2 (Cheese)
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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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~Mathkiddie
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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.
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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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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 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\pmod{4}</cmath><cmath>3A\equiv0\pmod{4}</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>.
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~Technodoggo ~zgahzlkw (small edits) ~ESAOPS (LaTeX)
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Solution 4
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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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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
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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> ~hollph27 ~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 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 (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.
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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
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Video Solution
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https://youtu.be/5OuzPFvJPEY
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Video Solution
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https://www.youtube.com/watch?v=Z-OCnHUwnj0
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Video Solution by OmegaLearn
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https://youtu.be/0AG5XmWY-D8
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Video Solution by TheBeautyofMath
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https://www.youtube.com/watch?v=zvKijDeiYUs
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Video Solution
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https://youtu.be/0ssjr8KjOzk
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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See Also Cheese
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2023 AMC 10A (Problems • Answer Key • Resources)
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Preceded by
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Problem 17 Followed by
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Problem 19
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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
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All AMC 10 Problems and Solutions
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Latest revision as of 13:40, 14 October 2024

-Note: this page was griefed by Alac 16 and I am trying my best to restore it. Formatting help would be much appreciated-Toucan2009hz Contents 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 (Dual) 9 Video Solution by Math-X (First fully understand the problem!!!) 10 Video Solution 11 Video Solution 12 Video Solution by OmegaLearn 13 Video Solution by TheBeautyofMath 14 Video Solution 15 See Also Cheese 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.

\[3A+4B\equiv48\pmod{4}\]\[3A\equiv0\pmod{4}\]

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 (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 The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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