2023 AMC 8 Problems/Problem 17: Difference between revisions
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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 <math>Q</math>? | 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 <math>Q</math>? | ||
<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> | |||
==Solution (Intuition)== | ==Solution (Intuition)== | ||
Revision as of 13:50, 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 
?
![[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
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
