Difference between revisions of "2023 AMC 10A Problems/Problem 18"
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==Solution 2== | ==Solution 2== | ||
− | With 12 rhombi, there are <math>48</math> | + | With <math>12</math> rhombi, there are <math>4\cdot12=48</math> total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have <math>\dfrac{48}2=24</math> total edges. |
− | Let <math>A</math> be the number of | + | Let <math>A</math> be the number of vertices with <math>3</math> edges (this is what the problem asks for) and <math>B</math> be the number of vertices with <math>4</math> edges. We have <math>3A + 4B = 48</math>. |
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− | |||
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− | + | Euler's formula states that, for all convex polyhedra, <math>v-e+f=2</math>. In our case, <math>v-24+12=2\implies v=14.</math> We know that <math>A+B</math> is the total number of vertices as we are given that all vertices are connected to either <math>3</math> or <math>4</math> edges. Therefore, <math>A+B=14.</math> | |
− | + | ||
+ | We now have a system of two equations. There are many ways to solve for <math>A</math>; choosing one yields <math>A=\boxed}\textbf{(D) }8</math>. | ||
+ | |||
+ | Even without Euler's formula, we can do a bit of answer guessing. From <math>3A+4B=48</math>, we take mod <math>4</math> on both sides. | ||
+ | |||
+ | <cmath>3A+4B\equiv48~(\mod4)</cmath> | ||
+ | <cmath>3A\equiv0~(\mod4)</cmath> | ||
+ | |||
+ | We know that <math>3A</math> must be divisible by <math>4</math>. We know that the factor of <math>3</math> will not affect the divisibility by <math>4</math> of <math>3A</math>, so we remove the <math>3</math>. We know that <math>A</math> is divisible by <math>4</math>. Checking answer choices, the only one divisible by <math>4</math> is indeed <math>A=\boxed}\textbf{(D) }8</math>. | ||
~Technodoggo ~zgahzlkw (small edits) | ~Technodoggo ~zgahzlkw (small edits) |
Revision as of 22:08, 9 November 2023
Problem
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 1
Note Euler's formula where . There are
faces and the number of edges is
because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are
vertices on the figure. Let
be the number of vertices with degree 3 and
be the number of vertices with degree 4.
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
. Solving this system of equations gives
and
so the answer is
.
~aiden22gao ~zgahzlkw (LaTeX)
Solution 2
With rhombi, there are
total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have
total edges.
Let be the number of vertices with
edges (this is what the problem asks for) and
be the number of vertices with
edges. We have
.
Euler's formula states that, for all convex polyhedra, . In our case,
We know that
is the total number of vertices as we are given that all vertices are connected to either
or
edges. Therefore,
We now have a system of two equations. There are many ways to solve for ; choosing one yields $A=\boxed}\textbf{(D) }8$ (Error compiling LaTeX. Unknown error_msg).
Even without Euler's formula, we can do a bit of answer guessing. From , we take mod
on both sides.
We know that must be divisible by
. We know that the factor of
will not affect the divisibility by
of
, so we remove the
. We know that
is divisible by
. Checking answer choices, the only one divisible by
is indeed $A=\boxed}\textbf{(D) }8$ (Error compiling LaTeX. Unknown error_msg).
~Technodoggo ~zgahzlkw (small edits)
Solution 3
Note that Euler's formula is . We know
from the question. We also know
because every face has
edges and every edge is shared by
faces. We can solve for the vertices based on this information.
Using the formula we can find:
Let
be the number of vertices with
edges and
be the number of vertices with
edges. We know
from the question and
. The second equation is because the total number of points is
because there are 12 rhombuses of
vertices.
Now, we just have to solve a system of equations.
Our answer is simply just
, which is
~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 -point intersections, we have a grid of squares. If both occur at
-point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a
-point intersection and two at a
-point intersection.
Since each -point intersection has
adjacent rhombuses, we know the number of
-point intersections must equal the number of
-point intersections per rhombus times the number of rhombuses over
. Since there are
rhombuses and two
-point intersections per rhombus, this works out to be:
Hence:
~hollph27
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.