Difference between revisions of "2023 AMC 10A Problems/Problem 18"

Revision as of 20:37, 9 November 2023

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

Solution 4

Each of the twelve rhombuses has two pairs of angles across from each other that must be the same. If both pairs of angles occur at $4$ -point intersections, we have a grid of squares. If both occur at $3$ -point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a $3$ -point intersection and two at a $4$ -point intersection.

Since each $3$ -point intersection has $3$ adjacent rhombuses, we know the number of $3$ -point intersections must equal the number of $3$ -point intersections per rhombus times the number of rhombuses over $3$ . Since there are $12$ rhombuses and two $3$ -point intersections per rhombus, this works out to be:

\frac{2*12}{3}

Hence: $\fbox{(D) 8}$ ~hollph27

@@ Line 47: / Line 47: @@
 Hence: <math>\fbox{(D) 8}</math>
 ~hollph27
+==See Also==
+{{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}}
+{{MAA Notice}}

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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