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

Revision as of 21:18, 25 January 2023

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 $Q$ ?

$[asy] // Diagram by TheMathGuyd import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*"$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 = (6.92820323028,0); DB = (3.46410161514,6); 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 (Intuition)

The answer is $\boxed{\textbf{(A)}\ 1}.$ Use intuition to bring it down to $2$ guesses $1$ or $2$ and guess from there or you could actually fold the paper. -apex304

Solution 2

In the net, face $5$ shares a vertex with the bottom-left corner of face $Q$ , but it doesn't share an edge with face $Q$ , so face $5$ must be the face diagonally across from face $Q$ 's bottom-left corner in the octahedron. Notice that in the octahedron, this face is the only one that doesn't share any vertices with face $?$ . On the net, face $5$ shares at least 1 vertex with all of the other faces except for face $1$ . Thus, the number of face $?$ is $\boxed{\textbf{(A)}\ 1}$ . -UnknownMonkey

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

@@ Line 54: / Line 54: @@
 ==Solution 2==
-In the net, face <math>5</math> shares a vertex with the bottom-left corner of face <math>Q</math>, but it doesn't share an edge with face <math>Q</math>, so face <math>5</math> must be the face diagonally across from face <math>Q</math>'s bottom-left corner in the octahedron. Notice that in the octahedron, this face is the only one that doesn't share any vertices with face <math>?</math>. On the net, face <math>5</math> shares at least 1 vertex with all of the other faces except for face <math>1</math>. Thus, the number of face <math>?</math> is <math>\boxed{\text{(A)}~1}</math>. -UnknownMonkey
+In the net, face <math>5</math> shares a vertex with the bottom-left corner of face <math>Q</math>, but it doesn't share an edge with face <math>Q</math>, so face <math>5</math> must be the face diagonally across from face <math>Q</math>'s bottom-left corner in the octahedron. Notice that in the octahedron, this face is the only one that doesn't share any vertices with face <math>?</math>. On the net, face <math>5</math> shares at least 1 vertex with all of the other faces except for face <math>1</math>. Thus, the number of face <math>?</math> is <math>\boxed{\textbf{(A)}\ 1}</math>. -UnknownMonkey
 ==Animated Video Solution==

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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