Difference between revisions of "2023 AMC 8 Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | A | + | A <i>regular octahedron</i> has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of <math>Q</math>? |
− | + | <asy> | |
+ | // Diagram by TheMathGuyd | ||
+ | import graph; | ||
+ | // The Solid | ||
+ | // To save processing time, do not use three (dimensions) | ||
+ | // Project (roughly) to two | ||
+ | size(15cm); | ||
+ | pair Fr, Lf, Rt, Tp, Bt, Bk; | ||
+ | Lf=(0,0); | ||
+ | Rt=(12,1); | ||
+ | Fr=(7,-1); | ||
+ | Bk=(5,2); | ||
+ | Tp=(6,6.7); | ||
+ | Bt=(6,-5.2); | ||
+ | draw(Lf--Fr--Rt); | ||
+ | draw(Lf--Tp--Rt); | ||
+ | draw(Lf--Bt--Rt); | ||
+ | draw(Tp--Fr--Bt); | ||
+ | draw(Lf--Bk--Rt,dashed); | ||
+ | draw(Tp--Bk--Bt,dashed); | ||
+ | label(rotate(-8.13010235)*slant(0.1)*"$Q$", (4.2,1.6)); | ||
+ | label(rotate(21.8014095)*slant(-0.2)*"$?$", (8.5,2.05)); | ||
+ | pair g = (-8,0); // Define Gap transform | ||
+ | real a = 8; | ||
+ | draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow | ||
+ | // Time for the NET | ||
+ | pair DA,DB,DC,CD,O; | ||
+ | DA = (4*sqrt(3),0); | ||
+ | DB = (2*sqrt(3),6); | ||
+ | DC = (DA+DB)/3; | ||
+ | CD = conj(DC); | ||
+ | O=(0,0); | ||
+ | transform trf=shift(3g+(0,3)); | ||
+ | path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); | ||
+ | draw(trf*NET); | ||
+ | label("$7$",trf*DC); | ||
+ | label("$Q$",trf*DC+DA-DB); | ||
+ | label("$5$",trf*DC-DB); | ||
+ | label("$3$",trf*DC-DA-DB); | ||
+ | label("$6$",trf*CD); | ||
+ | label("$4$",trf*CD-DA); | ||
+ | label("$2$",trf*CD-DA-DB); | ||
+ | label("$1$",trf*CD-2DA); | ||
+ | </asy> | ||
+ | |||
+ | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | We color face <math>6</math> red and face <math>5</math> yellow. Note that from the octahedron, face <math>5</math> and face <math>?</math> do not share anything in common. From the net, face <math>5</math> shares at least one vertex with all other faces except face <math>1,</math> which is shown in green: | ||
+ | <asy> | ||
+ | /* | ||
+ | Diagram by TheMathGuyd | ||
+ | Edited by MRENTHUSIASM | ||
+ | */ | ||
+ | import graph; | ||
+ | // The Solid | ||
+ | // To save processing time, do not use three (dimensions) | ||
+ | // Project (roughly) to two | ||
+ | size(15cm); | ||
+ | pair Fr, Lf, Rt, Tp, Bt, Bk; | ||
+ | Lf=(0,0); | ||
+ | Rt=(12,1); | ||
+ | Fr=(7,-1); | ||
+ | Bk=(5,2); | ||
+ | Tp=(6,6.7); | ||
+ | Bt=(6,-5.2); | ||
+ | fill(Tp--Bk--Lf--cycle,red); | ||
+ | fill(Bt--Bk--Lf--cycle,yellow); | ||
+ | fill(Fr--Rt--Tp--cycle,green); | ||
+ | draw(Lf--Fr--Rt); | ||
+ | draw(Lf--Tp--Rt); | ||
+ | draw(Lf--Bt--Rt); | ||
+ | draw(Tp--Fr--Bt); | ||
+ | draw(Lf--Bk--Rt,dashed); | ||
+ | draw(Tp--Bk--Bt,dashed); | ||
+ | label(rotate(-8.13010235)*slant(0.1)*"$Q$", (4.2,1.6)); | ||
+ | label(rotate(21.8014095)*slant(-0.2)*"$?$", (8.5,2.05)); | ||
+ | pair g = (-8,0); // Define Gap transform | ||
+ | real a = 8; | ||
+ | draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow | ||
+ | // Time for the NET | ||
+ | pair DA,DB,DC,CD,O; | ||
+ | DA = (4*sqrt(3),0); | ||
+ | DB = (2*sqrt(3),6); | ||
+ | DC = (DA+DB)/3; | ||
+ | CD = conj(DC); | ||
+ | O=(0,0); | ||
+ | transform trf=shift(3g+(0,3)); | ||
+ | path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); | ||
+ | fill(trf*((DA-DB)--O--DA--cycle),red); | ||
+ | fill(trf*((DA-DB)--O--(-DB)--cycle),yellow); | ||
+ | fill(trf*((-2*DA)--(-DA-DB)--(-DA)--cycle),green); | ||
+ | draw(trf*NET); | ||
+ | label("$7$",trf*DC); | ||
+ | label("$Q$",trf*DC+DA-DB); | ||
+ | label("$5$",trf*DC-DB); | ||
+ | label("$3$",trf*DC-DA-DB); | ||
+ | label("$6$",trf*CD); | ||
+ | label("$4$",trf*CD-DA); | ||
+ | label("$2$",trf*CD-DA-DB); | ||
+ | label("$1$",trf*CD-2DA); | ||
+ | </asy> | ||
+ | Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}.</math> | ||
+ | |||
+ | ~UnknownMonkey, apex304, MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2== | ||
+ | We label the octohedron going triangle by triangle until we reach the <math>?</math> triangle. The triangle to the left of the <math>Q</math> should be labeled with a <math>6</math>. Underneath triangle <math>6</math> is triangle <math>5</math>. The triangle to the right of triangle <math>5</math> is triangle <math>4</math> and further to the right is triangle <math>3</math>. Finally, the side of triangle <math>3</math> under triangle <math>Q</math> is <math>2</math>, so the triangle to the right of <math>Q</math> is <math>\boxed{\textbf{(A)}\ 1}</math>. | ||
+ | |||
+ | ~hdanger | ||
+ | |||
+ | ==Solution 3 (Fast and Cheap)== | ||
+ | Notice that the triangles labeled <math>2, 3, 4,</math> and <math>5</math> make the bottom half of the octahedron, as shown below: | ||
+ | <asy> | ||
+ | /* | ||
+ | Diagram by TheMathGuyd | ||
+ | Edited by MRENTHUSIASM | ||
+ | */ | ||
+ | import graph; | ||
+ | // The Solid | ||
+ | // To save processing time, do not use three (dimensions) | ||
+ | // Project (roughly) to two | ||
+ | size(15cm); | ||
+ | pair Fr, Lf, Rt, Tp, Bt, Bk; | ||
+ | Lf=(0,0); | ||
+ | Rt=(12,1); | ||
+ | Fr=(7,-1); | ||
+ | Bk=(5,2); | ||
+ | Tp=(6,6.7); | ||
+ | Bt=(6,-5.2); | ||
+ | dot(Bt,linewidth(5)); | ||
+ | draw(Lf--Fr--Rt); | ||
+ | draw(Lf--Tp--Rt); | ||
+ | draw(Lf--Bt--Rt); | ||
+ | draw(Tp--Fr--Bt); | ||
+ | draw(Lf--Bk--Rt,dashed); | ||
+ | draw(Tp--Bk--Bt,dashed); | ||
+ | label(rotate(-8.13010235)*slant(0.1)*"$Q$", (4.2,1.6)); | ||
+ | label(rotate(21.8014095)*slant(-0.2)*"$?$", (8.5,2.05)); | ||
+ | pair g = (-8,0); // Define Gap transform | ||
+ | real a = 8; | ||
+ | draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow | ||
+ | // Time for the NET | ||
+ | pair DA,DB,DC,CD,O; | ||
+ | DA = (4*sqrt(3),0); | ||
+ | DB = (2*sqrt(3),6); | ||
+ | DC = (DA+DB)/3; | ||
+ | CD = conj(DC); | ||
+ | O=(0,0); | ||
+ | transform trf=shift(3g+(0,3)); | ||
+ | path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); | ||
+ | dot(trf*(-DB),linewidth(5)); | ||
+ | draw(trf*NET); | ||
+ | label("$7$",trf*DC); | ||
+ | label("$Q$",trf*DC+DA-DB); | ||
+ | label("$5$",trf*DC-DB); | ||
+ | label("$3$",trf*DC-DA-DB); | ||
+ | label("$6$",trf*CD); | ||
+ | label("$4$",trf*CD-DA); | ||
+ | label("$2$",trf*CD-DA-DB); | ||
+ | label("$1$",trf*CD-2DA); | ||
+ | </asy> | ||
+ | Therefore, <math>\textbf{(B)}, \textbf{(C)}, \textbf{(D)},</math> and <math>\textbf{(E)}</math> are clearly not the correct answer. Thus, the only choice left is <math>\boxed{\textbf{(A)}\ 1}</math>. | ||
+ | |||
+ | ~andy_lee | ||
+ | |||
+ | ==Solution 4 (Really Simple Reasoning)== | ||
+ | |||
+ | The first half of the octahedron will need <math>4</math> triangles connected to one another to form it. We can choose the triangles <math>4</math>, <math>5</math>, <math>6</math>, and <math>7</math> and form the half around the vertex they all share. That leaves triangles <math>1</math>, <math>3</math>, <math>2</math>, and <math>Q</math> to form the second half. Triangle <math>3</math> will definitely share its sides with triangles <math>1</math> and <math>2</math>, leaving them to share their second side with triangle <math>Q</math>. Since triangle <math>Q</math> will certainly share its left side with triangle <math>2</math>, the only triangle left to share its right side is triangle <math>\boxed{\textbf{(A)}\ 1}</math> | ||
+ | |||
+ | ~mihikamishra | ||
+ | |||
+ | ==Video Solution by Math-X (Simple Visualization)== | ||
+ | https://youtu.be/Ku_c1YHnLt0?si=XilaQFDcnGR8ak7W&t=3364 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/tcGcqm5RHiY | ||
+ | |||
+ | ~Education, the Study of Everything | ||
− | |||
− | |||
− | |||
==Animated Video Solution== | ==Animated Video Solution== | ||
https://youtu.be/ECqljkDeA5o | https://youtu.be/ECqljkDeA5o | ||
Line 13: | Line 190: | ||
==Video Solution by OmegaLearn (Using 3D Visualization)== | ==Video Solution by OmegaLearn (Using 3D Visualization)== | ||
− | https:// | + | https://www.youtube.com/watch?v=nVSF2ujZkPE&ab_channel=SohilRathi |
==Video Solution by Magic Square== | ==Video Solution by Magic Square== | ||
https://youtu.be/-N46BeEKaCQ?t=3789 | https://youtu.be/-N46BeEKaCQ?t=3789 | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/DBqko2xATxs&t=2195 | ||
+ | |||
+ | ==Video Solution (Solve under 60 seconds!!!)== | ||
+ | https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=786 | ||
+ | |||
+ | ~hsnacademy | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=16|num-a=18}} | {{AMC8 box|year=2023|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 02:06, 21 August 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Fast and Cheap)
- 5 Solution 4 (Really Simple Reasoning)
- 6 Video Solution by Math-X (Simple Visualization)
- 7 Video Solution (CREATIVE THINKING!!!)
- 8 Animated Video Solution
- 9 Video Solution by OmegaLearn (Using 3D Visualization)
- 10 Video Solution by Magic Square
- 11 Video Solution by Interstigation
- 12 Video Solution (Solve under 60 seconds!!!)
- 13 See Also
Problem
A regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of ?
Solution 1
We color face red and face
yellow. Note that from the octahedron, face
and face
do not share anything in common. From the net, face
shares at least one vertex with all other faces except face
which is shown in green:
Therefore, the answer is
~UnknownMonkey, apex304, MRENTHUSIASM
Solution 2
We label the octohedron going triangle by triangle until we reach the triangle. The triangle to the left of the
should be labeled with a
. Underneath triangle
is triangle
. The triangle to the right of triangle
is triangle
and further to the right is triangle
. Finally, the side of triangle
under triangle
is
, so the triangle to the right of
is
.
~hdanger
Solution 3 (Fast and Cheap)
Notice that the triangles labeled and
make the bottom half of the octahedron, as shown below:
Therefore,
and
are clearly not the correct answer. Thus, the only choice left is
.
~andy_lee
Solution 4 (Really Simple Reasoning)
The first half of the octahedron will need triangles connected to one another to form it. We can choose the triangles
,
,
, and
and form the half around the vertex they all share. That leaves triangles
,
,
, and
to form the second half. Triangle
will definitely share its sides with triangles
and
, leaving them to share their second side with triangle
. Since triangle
will certainly share its left side with triangle
, the only triangle left to share its right side is triangle
~mihikamishra
Video Solution by Math-X (Simple Visualization)
https://youtu.be/Ku_c1YHnLt0?si=XilaQFDcnGR8ak7W&t=3364
~Math-X
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Using 3D Visualization)
https://www.youtube.com/watch?v=nVSF2ujZkPE&ab_channel=SohilRathi
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=3789
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=2195
Video Solution (Solve under 60 seconds!!!)
https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=786
~hsnacademy
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.