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| − | ==Problem==
| + | User:AlacCary_466 |
| − | 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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| − | <math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math>
| + | Gmail: alaccary446@gmail.com |
| − | | + | Hashtag:#goole trends #google news |
| − | ==Solution 1==
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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 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>
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| − | <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:
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| − | <cmath>V + 12 - 24 = 2</cmath>
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| − | <cmath>V = 14</cmath>
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| − | 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.
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| − | Now, we just have to solve a system of equations.
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| − | <cmath>3t + 4f = 48</cmath>
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| − | <cmath>3t + 3f = 42</cmath>
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| − | <cmath>f = 6</cmath>
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| − | <cmath>t = 8</cmath>
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| − | Our answer is simply just <math>t</math>, which is <math>\boxed{\textbf{(D) }8}</math>
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| − | ~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 (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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| − | {{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}}
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| − | {{MAA Notice}}
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