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

Difference between revisions of "2022 AMC 8 Problems/Problem 24"

(Since the two solutions are very similar. I will make Solution 2 supplement of Solution 1.)
m
Line 32: Line 32:
 
While imagining the folding, <math>\overline{AB}</math> goes on <math>\overline{BC},</math> <math>\overline{AH}</math> goes on <math>\overline{CI},</math> and <math>\overline{EF}</math> goes on <math>\overline{FG}.</math> So, <math>\overline{BJ}=\overline{CI}=8</math> and <math>\overline{FG}=\overline{BC}=8.</math> Also, <math>\overline{HJ}</math> becomes an edge parallel to <math>\overline{FG},</math> so that means <math>\overline{HJ}=8.</math>
 
While imagining the folding, <math>\overline{AB}</math> goes on <math>\overline{BC},</math> <math>\overline{AH}</math> goes on <math>\overline{CI},</math> and <math>\overline{EF}</math> goes on <math>\overline{FG}.</math> So, <math>\overline{BJ}=\overline{CI}=8</math> and <math>\overline{FG}=\overline{BC}=8.</math> Also, <math>\overline{HJ}</math> becomes an edge parallel to <math>\overline{FG},</math> so that means <math>\overline{HJ}=8.</math>
  
−
Since <math>\overline{GH}=14,</math> then <math>\overline{JG}=14-8=6.</math> So, the area of <math>\triangle BJG</math> is <math>\frac{8\cdot6}{2}=24.</math> If we let <math>\triangle BJG</math> be the base, then the height is <math>\overline{FG}=8.</math> So, the volume is <math>24\cdot8=\boxed{\textbf{(C) }192}.</math>
+
Since <math>\overline{GH}=14,</math> then <math>\overline{JG}=14-8=6.</math> So, the area of <math>\triangle BJG</math> is <math>\frac{8\cdot6}{2}=24.</math> If we let <math>\triangle BJG</math> be the base, then the height is <math>\overline{FG}=8.</math> So, the volume is <math>24\cdot8=\boxed{\textbf{(C)} ~192}.</math>
  
 
After folding polygon <math>ABCDEFGH</math> on the dotted lines, we obtain the following triangular prism:
 
After folding polygon <math>ABCDEFGH</math> on the dotted lines, we obtain the following triangular prism:

Revision as of 13:10, 10 June 2022

Problem

The figure below shows a polygon $ABCDEFGH$, consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that $AH = EF = 8$ and $GH = 14$. What is the volume of the prism?

[asy] usepackage("mathptmx"); size(275); defaultpen(linewidth(0.8)); real r = 2, s = 2.5, theta = 14; pair G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta); pair N = (B+G)/2, J = N + s/2 * dir(180+theta); pair E = F + r * dir(- 45 - theta/2), D = I+E-F; pair H = J + r * dir(135 + theta/2), A = B+H-J; draw(A--B--C--I--D--E--F--G--J--H--cycle^^rightanglemark(F,I,C)^^rightanglemark(G,J,B)); draw(J--B--G^^C--F--I,linetype ("4 4")); dot("$A$",A,N); dot("$B$",B,1.2*N); dot("$C$",C,N); dot("$D$",D,dir(0)); dot("$E$",E,S); dot("$F$",F,1.5*dir(-100)); dot("$G$",G,S); dot("$H$",H,W); dot("$I$",I,NE); dot("$J$",J,1.5*S); [/asy]

$\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288$

Solution

While imagining the folding, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $\overline{BJ}=\overline{CI}=8$ and $\overline{FG}=\overline{BC}=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $\overline{HJ}=8.$

Since $\overline{GH}=14,$ then $\overline{JG}=14-8=6.$ So, the area of $\triangle BJG$ is $\frac{8\cdot6}{2}=24.$ If we let $\triangle BJG$ be the base, then the height is $\overline{FG}=8.$ So, the volume is $24\cdot8=\boxed{\textbf{(C)} ~192}.$

After folding polygon $ABCDEFGH$ on the dotted lines, we obtain the following triangular prism: [asy] /* Made by MRENTHUSIASM */ usepackage("mathptmx"); size(200); defaultpen(linewidth(0.8)); import graph3; import solids; currentprojection=orthographic((0.3,-0.3,0.3)); triple J, G, B, A, H, F; J = (0,0,0); G = (6,0,0); B = (0,8,0); A = (0,8,8); H = (0,0,8); F = (6,0,8); draw(surface(B--J--G--cycle),yellow); draw(surface(H--A--F--cycle),yellow); draw(B--J,dashed); draw(G--J--H--A--B^^A--F--H^^F--G^^B--G); draw((0.5,0,0)--(0.5,0.5,0)--(0,0.5,0)^^(0.5,0,8)--(0.5,0.5,8)--(0,0.5,8)); dot("$A=C$",A,1.5*E); dot("$B$",B,1.5*E); dot("$D=J$",J,1.5*W); dot("$F$",F,1.5*E); dot("$H=I$",H,1.5*W); dot("$E=G$",G,1.5*E); label("$8$",midpoint(A--H),1.5*NW,red); label("$6$",midpoint(H--F),1.5*S,red); label("$8$",midpoint(H--J),1.5*W,red); [/asy] ~aops-g5-gethsemanea2 ~MRENTHUSIASM

Video Solution

https://www.youtube.com/watch?v=2uoBPp4Kxck

~Mathematical Dexterity

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=2432

~Interstigation

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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=2022_AMC_8_Problems/Problem_24&oldid=174624"