Difference between revisions of "2023 AMC 8 Problems/Problem 17"
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==Solution 1== | ==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> as shown in green. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}.</math> | + | 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> as shown in green. |
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| − | + | Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}.</math> | |
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~UnknownMonkey, apex304, MRENTHUSIASM | ~UnknownMonkey, apex304, MRENTHUSIASM | ||
Revision as of 03:08, 27 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 = (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]](http://latex.artofproblemsolving.com/7/3/5/73573481f96dd080a6118ceb2a3c4588148704dc.png)
Solution 1
We color face 
red and face 
yellow. Note that from 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 
~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. 
