Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1986 AJHSME Problems/Problem 21

Revision as of 14:18, 28 September 2020 by Bakedpotato66 (talk | contribs) (→‎Solution 1)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a topless cubical box?

[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cycle); draw((1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--cycle,linewidth(.7 mm)); label("A",(1.5,4.2),N); label("B",(.5,3.2),N); label("C",(2.5,3.2),N); label("D",(.5,2.2),N); label("E",(3.5,2.2),N); label("F",(.5,1.2),N); label("G",(2.5,1.2),N); label("H",(1.5,.2),N); [/asy]

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution

Solution 1

The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of these two places.

  • $A$ would be the top side - $\text{OK}$
  • $B$ would be the left side - $\text{OK}$
  • $C$ would cause the figure to not be foldable at all - NO
  • $D$ would be the left side - $\text{OK}$
  • $E$ would be the top side - $\text{OK}$
  • $F$ is the same case as $B$ - $\text{OK}$
  • $G$ is the same case as $C$ - NO
  • $H$ is the same case as $A$ - $\text{OK}$

In total, there are $6\rightarrow\boxed{\text{E}}$ good possibilities.

Solution 2

Fold the four squares into the four sides of a cube. Then, there are six edges "open" (for lack of better term). For each open edge, we can add a square/side, so the answer is $6\rightarrow\boxed{\text{E}}$.

See Also

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_21&oldid=134221"