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2023 AMC 10A Problems/Problem 18

Revision as of 20:11, 9 November 2023 by Zgahzlkw (talk | contribs) (→‎Problem)

Problem

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 1

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 $\fbox{D}$. ~aiden22gao ~zgahzlkw (LaTeX)

Solution 2

With 12 rhombi, there are $48$ sides. All the sides are shared by 2 faces. Thus we have $24$ shared sides/edges.

Let $A$ be the number of edges with 3 vertices and $B$ be the number of edges with 4 vertices. We get $3A + 4B = 48$. With Euler's formula, $V-3+F=2$. $V-24+12=2$, so $V = 14$. Thus, $a+b= 14$. Solving the 2 equations, we get $a = 8$ and $b = 6$.

Even without Euler's formula, we observe that a must be even integers, so trying even integer choices and we also get $a = 8$. 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.

~Technodoggo ~zgahzlkw (small edits)

Solution 3

Note that Euler's formula is $V+F-E=2$. We know $F=12$ from the question. We also know $E = \frac{12 \cdot 4}{2} = 24$ because every face has $4$ edges and every edge is shared by $2$ faces. We can solve for the vertices based on this information.

Using the formula we can find: \[V + 12 - 24 = 2\] \[V = 14\] Let $t$ be the number of vertices with $3$ edges and $f$ be the number of vertices with $4$ edges. We know $t+f = 14$ from the question and $3t + 4f = 48$. The second equation is because the total number of points is $48$ because there are 12 rhombuses of $4$ vertices. Now, we just have to solve a system of equations. \[3t + 4f = 48\] \[3t + 3f = 42\] \[f = 6\] \[t = 8\] Our answer is simply just $t$, which is $\fbox{(D) 8}$ ~musicalpenguin

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