Difference between revisions of "2023 AMC 8 Problems/Problem 17"
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~hdanger | ~hdanger | ||
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+ | ==Solution 3 (Fast, Cheap)== | ||
+ | Notice that the triangles labeled <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> make one-half of the octahedron, so they are clearly not the correct answer. Thus, the only choice left is <math>\boxed{\textbf{(A)}\ 1}</math>. | ||
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+ | ~andy_lee | ||
==Animated Video Solution== | ==Animated Video Solution== |
Revision as of 11:08, 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 ?
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
Solution 3 (Fast, Cheap)
Notice that the triangles labeled , , , and make one-half of the octahedron, so they are clearly not the correct answer. Thus, the only choice left is .
~andy_lee
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.