Difference between revisions of "2023 AMC 10A Problems/Problem 18"
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==Solution 1== | ==Solution 1== | ||
− | Note Euler's formula where <math> | + | Note Euler's formula where <math>\text{Vertices}+\text{Faces}-\text{Edges}=2</math>. |
− | ~aiden22gao ~zgahzlkw (LaTeX) | + | There are <math>12</math> faces. |
+ | There are <math>24</math> edges, because there are 12 faces each with four edges and each edge is shared by two faces. | ||
+ | Now we know that there are <math>2-12+24=14</math> vertices. | ||
+ | Now note that the sum of the degrees of all the points is <math>2</math>(the number of edges). Let <math>x=</math> the number of vertices with <math>3</math> edges. Now we know <math>\frac{3x+4(14-x)}{2}=24</math>. Solving this equation gives <math>x = \boxed{\textbf{(D) }8}</math>. | ||
+ | ~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified) | ||
− | ==Solution 2== | + | ==Solution 2 (Cheese)== |
− | + | Let <math>x</math> be the number of vertices with 3 edges, and <math>y</math> be the number of vertices with 4 edges. Since there are <math>\frac{4*12}{2}=24</math> edges on the polyhedron, we can see that <math>\frac{3x+4y}{2}=24</math>. Then, <math>3x+4y=48</math>. Notice that by testing the answer choices, (D) is the only one that yields an integer solution for <math>y</math>. Thus, the answer is <math>\boxed{\textbf{(D) }8}</math>. | |
− | + | ~Mathkiddie | |
− | |||
− | |||
− | |||
− | + | ==Solution 3== | |
− | + | ||
+ | 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 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>. | ||
+ | |||
+ | 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~(\mod{4})</cmath> | ||
+ | <cmath>3A\equiv0~(\mod{4})</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>. | |
− | ==Solution | + | ~Technodoggo ~zgahzlkw (small edits) ~ESAOPS (LaTeX) |
+ | |||
+ | ==Solution 4== | ||
Note that Euler's formula is <math>V+F-E=2</math>. We know <math>F=12</math> from the question. We also know <math>E = \frac{12 \cdot 4}{2} = 24</math> because every face has <math>4</math> edges and every edge is shared by <math>2</math> faces. We can solve for the vertices based on this information. | Note that Euler's formula is <math>V+F-E=2</math>. We know <math>F=12</math> from the question. We also know <math>E = \frac{12 \cdot 4}{2} = 24</math> because every face has <math>4</math> edges and every edge is shared by <math>2</math> faces. We can solve for the vertices based on this information. | ||
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<cmath>f = 6</cmath> | <cmath>f = 6</cmath> | ||
<cmath>t = 8</cmath> | <cmath>t = 8</cmath> | ||
− | Our answer is simply just <math>t</math>, which is <math>\ | + | Our answer is simply just <math>t</math>, which is <math>\boxed{\textbf{(D) }8}</math> |
~musicalpenguin | ~musicalpenguin | ||
− | ==Solution | + | ==Solution 5== |
− | Each of the twelve | + | Each of the twelve rhombi has two pairs of angles across from each other that must be congruent. If both pairs of angles occur at <math>4</math>-point intersections, we have a grid of squares. If both occur at <math>3</math>-point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a <math>3</math>-point intersection and two at a <math>4</math>-point intersection. |
Since each <math>3</math>-point intersection has <math>3</math> adjacent rhombuses, we know the number of <math>3</math>-point intersections must equal the number of <math>3</math>-point intersections per rhombus times the number of rhombuses over <math>3</math>. Since there are <math>12</math> rhombuses and two <math>3</math>-point intersections per rhombus, this works out to be: | Since each <math>3</math>-point intersection has <math>3</math> adjacent rhombuses, we know the number of <math>3</math>-point intersections must equal the number of <math>3</math>-point intersections per rhombus times the number of rhombuses over <math>3</math>. Since there are <math>12</math> rhombuses and two <math>3</math>-point intersections per rhombus, this works out to be: | ||
− | \frac{2 | + | <math>\frac{2\cdot12}{3}</math> |
− | Hence: <math>\ | + | Hence: <math>\boxed{\textbf{(D) }8}</math> |
~hollph27 | ~hollph27 | ||
+ | ~Minor edits by FutureSphinx | ||
+ | |||
+ | ==Solution 6 (Based on previous knowledge)== | ||
+ | Note that a rhombic dodecahedron is formed when a cube is turned inside out (as seen [https://en.wikipedia.org/wiki/Rhombic_dodecahedron#/media/File:R1-cube.gif here]), thus there are 6 4-vertices (corresponding to each face of the cube) and 8 3-vertices (corresponding to each corner of the cube). Thus the answer is <math>\boxed{\textbf{(D) }8}</math> | ||
+ | |||
+ | ==Solution 7 (Using Answer Choices)== | ||
+ | Let <math>m</math> be the number of <math>4</math>-edge vertices, and <math>n</math> be the number of <math>3</math>-edge vertices. The total number of vertices is <math>m+n</math>. Now, we know that there are <math>4 \cdot 12 = 48</math> vertices, but we have overcounted. We have overcounted <math>m</math> vertices <math>3</math> times and overcounted <math>n</math> vertices <math>2</math> times. Therefore, we subtract <math>3m</math> and <math>2n</math> from <math>48</math> and set it equal to our original number of vertices. | ||
+ | <cmath>48 - 3m - 2n = m+n</cmath> | ||
+ | <cmath>4m + 3n = 48</cmath> | ||
+ | From here, we reduce both sides modulo <math>4</math>. The <math>4m</math> disappears, and the left hand side becomes <math>3n</math>. The right hand side is <math>0</math>, meaning that <math>3n</math> must be divisible by <math>4</math>. Looking at the answer choices, this is only possible for <math>n = \boxed{8}</math>. | ||
+ | |||
+ | -DEVSAXENA | ||
+ | |||
+ | (Isn't this the same as the last half of Solution 2?) | ||
+ | |||
+ | ==Solution 8 (Dual)== | ||
+ | Note that a rhombic dodecahedron is the dual of a cuboctahedron. A cuboctahedron has <math>8</math> triangular faces, which correspond to <math>\boxed{\textbf{(D) }8}</math> vertices on a rhombic dodecahedron that have <math>3</math> edges. | ||
+ | |||
+ | ==Video Solution == | ||
+ | https://youtu.be/5OuzPFvJPEY | ||
+ | |||
+ | == Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=Z-OCnHUwnj0 | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/0AG5XmWY-D8 | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://www.youtube.com/watch?v=zvKijDeiYUs | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/0ssjr8KjOzk | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}} | {{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:50, 6 August 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Cheese)
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6 (Based on previous knowledge)
- 8 Solution 7 (Using Answer Choices)
- 9 Solution 8 (Dual)
- 10 Video Solution
- 11 Video Solution
- 12 Video Solution by OmegaLearn
- 13 Video Solution by TheBeautyofMath
- 14 Video Solution
- 15 See Also
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.
There are
edges, because there are 12 faces each with four edges and each edge is shared by two faces.
Now we know that there are
vertices.
Now note that the sum of the degrees of all the points is
(the number of edges). Let
the number of vertices with
edges. Now we know
. Solving this equation gives
.
~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)
Solution 2 (Cheese)
Let be the number of vertices with 3 edges, and
be the number of vertices with 4 edges. Since there are
edges on the polyhedron, we can see that
. Then,
. Notice that by testing the answer choices, (D) is the only one that yields an integer solution for
. Thus, the answer is
.
~Mathkiddie
Solution 3
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
.
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
.
~Technodoggo ~zgahzlkw (small edits) ~ESAOPS (LaTeX)
Solution 4
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 5
Each of the twelve rhombi has two pairs of angles across from each other that must be congruent. 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
~Minor edits by FutureSphinx
Solution 6 (Based on previous knowledge)
Note that a rhombic dodecahedron is formed when a cube is turned inside out (as seen here), thus there are 6 4-vertices (corresponding to each face of the cube) and 8 3-vertices (corresponding to each corner of the cube). Thus the answer is
Solution 7 (Using Answer Choices)
Let be the number of
-edge vertices, and
be the number of
-edge vertices. The total number of vertices is
. Now, we know that there are
vertices, but we have overcounted. We have overcounted
vertices
times and overcounted
vertices
times. Therefore, we subtract
and
from
and set it equal to our original number of vertices.
From here, we reduce both sides modulo
. The
disappears, and the left hand side becomes
. The right hand side is
, meaning that
must be divisible by
. Looking at the answer choices, this is only possible for
.
-DEVSAXENA
(Isn't this the same as the last half of Solution 2?)
Solution 8 (Dual)
Note that a rhombic dodecahedron is the dual of a cuboctahedron. A cuboctahedron has triangular faces, which correspond to
vertices on a rhombic dodecahedron that have
edges.
Video Solution
Video Solution
https://www.youtube.com/watch?v=Z-OCnHUwnj0
Video Solution by OmegaLearn
Video Solution by TheBeautyofMath
https://www.youtube.com/watch?v=zvKijDeiYUs
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.