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Difference between revisions of "2023 AMC 8 Problems/Problem 17"

(A carefully constructed Asymptote diagram has been added.)
 
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==Problem==
 
==Problem==
−
A ''regular octahedron'' has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedrons 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>?
+
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>
 
<asy>
Line 29: Line 29:
 
// Time for the NET
 
// Time for the NET
 
pair DA,DB,DC,CD,O;
 
pair DA,DB,DC,CD,O;
−
DA = (6.92820323028,0);
+
DA = (4*sqrt(3),0);
−
DB = (3.46410161514,6);
+
DB = (2*sqrt(3),6);
 
DC = (DA+DB)/3;
 
DC = (DA+DB)/3;
 
CD = conj(DC);
 
CD = conj(DC);
Line 47: Line 47:
 
</asy>
 
</asy>
  
−
==Solution (Intuition)==
+
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math>
−
The answer is <math>\boxed{\textbf{(A)}\ 1}.</math> Use intuition to bring it down to <math>2</math> guesses <math>1</math> or <math>2</math> and guess from there or you could actually fold the paper.
+
 
−
-apex304
+
==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 56: Line 190:
  
 
==Video Solution by OmegaLearn (Using 3D Visualization)==
 
==Video Solution by OmegaLearn (Using 3D Visualization)==
−
https://youtu.be/gIjhiw1CUgY
+
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
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/RKPKuD-fqLg
  
 
==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 23:55, 2 November 2024

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 $Q$?

[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]

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution 1

We color face $6$ red and face $5$ yellow. Note that from the octahedron, face $5$ and face $?$ do not share anything in common. From the net, face $5$ shares at least one vertex with all other faces except face $1,$ 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 $\boxed{\textbf{(A)}\ 1}.$

~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 $Q$ should be labeled with a $6$. Underneath triangle $6$ is triangle $5$. The triangle to the right of triangle $5$ is triangle $4$ and further to the right is triangle $3$. Finally, the side of triangle $3$ under triangle $Q$ is $2$, so the triangle to the right of $Q$ is $\boxed{\textbf{(A)}\ 1}$.

~hdanger

Solution 3 (Fast and Cheap)

Notice that the triangles labeled $2, 3, 4,$ and $5$ 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, $\textbf{(B)}, \textbf{(C)}, \textbf{(D)},$ and $\textbf{(E)}$ are clearly not the correct answer. Thus, the only choice left is $\boxed{\textbf{(A)}\ 1}$.

~andy_lee

Solution 4 (Really Simple Reasoning)

The first half of the octahedron will need $4$ triangles connected to one another to form it. We can choose the triangles $4$, $5$, $6$, and $7$ and form the half around the vertex they all share. That leaves triangles $1$, $3$, $2$, and $Q$ to form the second half. Triangle $3$ will definitely share its sides with triangles $1$ and $2$, leaving them to share their second side with triangle $Q$. Since triangle $Q$ will certainly share its left side with triangle $2$, the only triangle left to share its right side is triangle $\boxed{\textbf{(A)}\ 1}$

~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

https://youtu.be/ECqljkDeA5o

~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

Video Solution by Dr. David

https://youtu.be/RKPKuD-fqLg

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. AMC logo.png

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