Difference between revisions of "2023 AMC 10A Problems/Problem 17"
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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? | |
| + | ==Solution== | ||
| + | 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> | ||
| + | <cmath>3a+4b=48</cmath> | ||
| + | We find <math>b=6</math> and <math>a=8</math>, hence the answer is <math>8</math>. | ||
| + | |||
| + | ~SirAppel | ||
Revision as of 19:58, 9 November 2023
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
We see there are 
edges. We have by Euler's Polyhedral Formula, 
meaning 
or 
. Let there be 
vertices that have edges meeting and 
vertices that have edges meeting. Hence, ![\[a+b=14\]](http://latex.artofproblemsolving.com/d/d/0/dd0686d1c883a87048448c8954679ef5d5e86216.png)
![\[3a+4b=48\]](http://latex.artofproblemsolving.com/6/8/6/686c84861163a336a8ef4993a6e361db0ecb6540.png)
We find 
and 
, hence the answer is 
.
~SirAppel
