Difference between revisions of "2023 AMC 10A Problems/Problem 18"
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− | ==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>\text{Vertices}+\text{Faces}-\text{Edges}=2</math>. There are <math>12</math> faces | ||
− | ~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified) | ||
− | |||
− | |||
− | Let <math>x</math> be the number of vertices with | + | == Solution 2 == |
+ | 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 | ~Mathkiddie | ||
− | ==Solution 3== | + | == 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. | |
− | 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. | + | 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 | + | <cmath>3A+4B\equiv48\pmod{4}</cmath><cmath>3A\equiv0\pmod{4}</cmath> |
− | <cmath>3A\equiv0 | ||
− | 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) ~ESAOPS (LaTeX) | ~Technodoggo ~zgahzlkw (small edits) ~ESAOPS (LaTeX) | ||
− | ==Solution 4== | + | == 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. | + | 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 5== | + | == 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. | 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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<math>\frac{2\cdot12}{3}</math> | <math>\frac{2\cdot12}{3}</math> | ||
− | Hence: <math>\boxed{\textbf{(D) }8}</math> | + | Hence: <math>\boxed{\textbf{(D) }8}</math> ~hollph27 ~Minor edits by FutureSphinx |
− | ~hollph27 | ||
− | ~Minor edits by FutureSphinx | ||
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− | |||
− | |||
− | ==Solution | + | == 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> | |
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− | |||
− | |||
− | + | == 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 | + | == Video Solution == |
https://youtu.be/5OuzPFvJPEY | https://youtu.be/5OuzPFvJPEY | ||
− | == Video Solution | + | == Video Solution == |
− | |||
https://www.youtube.com/watch?v=Z-OCnHUwnj0 | https://www.youtube.com/watch?v=Z-OCnHUwnj0 | ||
− | ==Video Solution by OmegaLearn== | + | == Video Solution by OmegaLearn == |
https://youtu.be/0AG5XmWY-D8 | https://youtu.be/0AG5XmWY-D8 | ||
− | ==Video Solution by TheBeautyofMath== | + | == Video Solution by TheBeautyofMath == |
https://www.youtube.com/watch?v=zvKijDeiYUs | https://www.youtube.com/watch?v=zvKijDeiYUs | ||
− | ==Video Solution== | + | == Video Solution == |
− | |||
https://youtu.be/0ssjr8KjOzk | https://youtu.be/0ssjr8KjOzk | ||
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==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
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6 (Based on previous knowledge)
- 8 Solution 7 (Dual)
- 9 Video Solution by Little Fermat
- 10 Video Solution by Math-X (First fully understand the problem!!!)
- 11 Video Solution
- 12 Video Solution
- 13 Video Solution by OmegaLearn
- 14 Video Solution by TheBeautyofMath
- 15 Video Solution
- 16 See Also
Problem
A rhombic dodecahedron is a solid with congruent rhombus faces. At every vertex, or edges meet, depending on the vertex. How many vertices have exactly edges meet?
Solution 1
Note Euler's formula where . There are faces. There are edges, because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are vertices. Now note that the sum of the degrees of all the points is (the number of edges). Let the number of vertices with edges. Now we know . Solving this equation gives . ~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified) ~sonic12345 (Fixed typo)
Solution 2
Let be the number of vertices with 3 edges, and be the number of vertices with 4 edges. Since there are edges on the polyhedron, we can see that . Then, . Notice that by testing the answer choices, (D) is the only one that yields an integer solution for . Thus, the answer is .
~Mathkiddie
Solution 3
With rhombi, there are total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have total edges.
Let be the number of vertices with edges (this is what the problem asks for) and be the number of vertices with edges. We have .
Euler's formula states that, for all convex polyhedra, . In our case, We know that is the total number of vertices as we are given that all vertices are connected to either or edges. Therefore,
We now have a system of two equations. Solving the system yields .
Even without Euler's formula, we can do a bit of answer guessing. From , we take mod on both sides.
We know that must be divisible by . We know that the factor of will not affect the divisibility by of , so we remove the . We know that is divisible by . Checking answer choices, the only one divisible by is indeed .
~Technodoggo ~zgahzlkw (small edits) ~ESAOPS (LaTeX)
Solution 4
Note that Euler's formula is . We know from the question. We also know because every face has edges and every edge is shared by faces. We can solve for the vertices based on this information.
Using the formula we can find:Let be the number of vertices with edges and be the number of vertices with edges. We know from the question and . The second equation is because the total number of points is because there are 12 rhombuses of vertices. Now, we just have to solve a system of equations.Our answer is simply just , which is ~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 -point intersections, we have a grid of squares. If both occur at -point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a -point intersection and two at a -point intersection.
Since each -point intersection has adjacent rhombuses, we know the number of -point intersections must equal the number of -point intersections per rhombus times the number of rhombuses over . Since there are rhombuses and two -point intersections per rhombus, this works out to be:
Hence: ~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
Solution 7 (Dual)
Note that a rhombic dodecahedron is the dual of a cuboctahedron. A cuboctahedron has triangular faces, which correspond to vertices on a rhombic dodecahedron that have 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
Video Solution
https://www.youtube.com/watch?v=Z-OCnHUwnj0
Video Solution by OmegaLearn
Video Solution by TheBeautyofMath
https://www.youtube.com/watch?v=zvKijDeiYUs
Video Solution
~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.