2023 AMC 8 Problems/Problem 17
Contents
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 
?
![[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]](http://latex.artofproblemsolving.com/2/3/d/23d4f11f6be728f459c3ed7061962d0c9c38915d.png)
Solution (Intuition)
The answer is 
Use intuition to bring it down to 
guesses 
or and guess from there or you could actually fold the paper.
-apex304
Solution 2
In the net, face 
shares a vertex with the bottom-left corner of face , but it doesn't share an edge with face , so face must be the face diagonally across from face '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 shares at least 1 vertex with all of the other faces except for face . Thus, the number of face is \boxed{\text{(A)}~1}. -UnknownMonkey
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Using 3D Visualization)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
