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Difference between revisions of "2023 AMC 10A Problems/Problem 17"

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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>
 
<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math>
 
==Solution==
 
==Solution==
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We see there are <math>\frac{12 \cdot 4}{2}</math> edges. We have by Euler's Polyhedral Formula, <math>V-E+F=2</math> meaning <math>V-24+12=2</math> or <math>V=14</math>. Let there be <math>a</math> vertices that have <math>3</math> edges meeting and <math>b</math> vertices that have <math>4</math> edges meeting. Hence, <cmath>a+b=14</cmath>
 
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<cmath>3a+4b=48</cmath>
 
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We find <math>b=6</math> and <math>a=8</math>, hence the answer is <math>8</math>.
 
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~SirAppel
 

Revision as of 20:59, 9 November 2023

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

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