2023 AMC 8 Problems/Problem 17

Revision as of 14:50, 25 January 2023 by Themathguyd (talk | contribs) (A carefully constructed Asymptote diagram has been added.)

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]

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

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

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png