Difference between revisions of "2023 AMC 8 Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
| − | + | From the net, face <math>Q</math> and face <math>5</math> only share one vertex, namely the bottom-left vertex of face <math>Q.</math> So, face <math>5</math> must be diagonally across from face <math>Q</math>'s bottom-left vertex, as shown in red. Note that in 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> as shown in green. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}.</math> | |
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| + | DIAGRAM IN PROGRESS. | ||
~UnknownMonkey, apex304, MRENTHUSIASM | ~UnknownMonkey, apex304, MRENTHUSIASM | ||
Revision as of 16:48, 26 January 2023
Contents
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 
?
![[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 1
From the net, face and face 
only share one vertex, namely the bottom-left vertex of face 
So, face must be diagonally across from face 's bottom-left vertex, as shown in red. Note that in 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 
as shown in green. Therefore, the answer is 
DIAGRAM IN PROGRESS.
~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
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. 
