Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2: when you screw latex two times in a row)
(→‎Video Solution by Math-X (First fully understand the problem!!!))
 
(63 intermediate revisions by 28 users not shown)
Line 1: Line 1:
−
==Problem==
+
-Note: this page was griefed by Alac 16 and I am trying my best to restore it. Formatting help would be much appreciated-Toucan2009hz
 +
-fixed? wTaoTao
 +
 
 +
== 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?
 
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?
  
 
<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>
  
−
==Solution 1==
+
== 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>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)
  
−
Note Euler's formula where <math>V+F-E=2</math>. There are <math>12</math> faces and the number of edges is <math>24</math> because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are <math>14</math> vertices on the figure. Let <math>A</math> be the number of vertices with degree 3 and <math>B</math> be the number of vertices with degree 4. <math>A+B=14</math> 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 <math>3A+4B=48</math>. Solving this system of equations gives <math>B = 6</math> and <math>A = 8</math> so the answer is <math>\boxed{\textbf{(D) }8}</math>.
+
== Solution 2 ==
−
~aiden22gao ~zgahzlkw (LaTeX)
+
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>.
  
−
==Solution 2==
+
~Mathkiddie
  
−
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.  
+
== 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>.  
+
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>  
+
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>.  
+
We now have a system of two equations. Solving the system 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.  
+
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~(\mod4)</cmath>
+
<cmath>3A+4B\equiv48\pmod{4}</cmath><cmath>3A\equiv0\pmod{4}</cmath>
−
<cmath>3A\equiv0~(\mod4)</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>.  
+
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) ~ESAOPS (LaTeX)
  
−
==Solution 3==
+
== 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.
  
−
Using the formula we can find:  
+
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
−
<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 4==
+
== Solution 5 ==
−
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 <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.
+
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:
 
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*12}{3}</math>
+
<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 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 (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 Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=KLoAqmdqx_C55pEq&t=4063
 +
~little-fermat
 +
== 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
  
−
Hence: <math>\boxed{\textbf{(D) }8}</math>
+
== Video Solution ==
−
~hollph27
+
https://youtu.be/0ssjr8KjOzk
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See Also==
 
==See Also==
 +
https://en.wikipedia.org/wiki/Rhombic_dodecahedron
 +
 
{{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:34, 5 November 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 -fixed? wTaoTao

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

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. Solving the system 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 Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=KLoAqmdqx_C55pEq&t=4063 ~little-fermat

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

https://en.wikipedia.org/wiki/Rhombic_dodecahedron

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_18&oldid=231694"