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Difference between revisions of "2023 AMC 10A Problems/Problem 18"
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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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<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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someone else do latex pls
 
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Solution 1
 
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Note Euler's formula where V+F-E=2. There are 12 faces and the number of edges is 24 because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are 14 vertices on the figure. Let A be the number of vertices with degree 3 and B be the number of vertices with degree 4. A+B=14 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 3A+4B=48. Solving this system of equations gives B=6 and A = 8 so the answer is D.
 
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~aiden22gao
 
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Solution 2.
 
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With 12 rhombus, there are 12*4 sides. All the sides are shared by 2 faces. Thus we have 24 shared sides/ edges.
 
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Let a be the number of edges with 3 vertices and b be the number of edges with 4 verticies
 
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We got 3a + 4b = 48.
 
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With Euler's formula, v-3+f=2.  v-24+12=2, v=14. Thus, a+b= 14.
 
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Solving the 2 equations, we got a = 8, b = 12.
 
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Even without Euler's formula, we observe that a must be even integers, so trying even integer choices and we also get a = 8.
 
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Or with a keener number theory eye, we mod 4 on both side, leaving 3x mod 4 + 0 = 0.  Thus x, must be divisible by 4.
 
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~Technodoggo
 

Revision as of 07:38, 14 October 2024

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