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

(→‎Solution 3 (Fast, Cheap))
(→‎Solution 3 (Fast, Cheap))
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==Solution 3 (Fast, Cheap)==
 
==Solution 3 (Fast, 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:
 
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>.  
 
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>.  
  

Revision as of 19:13, 27 January 2023

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 $1,$ as 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, 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

Animated Video Solution

https://youtu.be/ECqljkDeA5o

~Star League (https://starleague.us)

Video Solution by OmegaLearn (Using 3D Visualization)

https://youtu.be/gIjhiw1CUgY

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=3789

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

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