Difference between revisions of "2022 AMC 8 Problems/Problem 24"
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Since <math>GH=14,</math> then <math>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>FG=8.</math> So, the volume is <math>24\cdot8=\boxed{\textbf{(C)} ~192}.</math> | Since <math>GH=14,</math> then <math>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>FG=8.</math> So, the volume is <math>24\cdot8=\boxed{\textbf{(C)} ~192}.</math> | ||
| + | ~aops-g5-gethsemanea2 | ||
| + | |||
| + | ==Remark== | ||
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: | ||
<asy> | <asy> | ||
| Line 66: | Line 69: | ||
label("$8$",midpoint(H--J),1.5*W,red); | label("$8$",midpoint(H--J),1.5*W,red); | ||
</asy> | </asy> | ||
| − | + | ~MRENTHUSIASM | |
==Video Solution== | ==Video Solution== | ||
Revision as of 19:34, 17 October 2022
Problem
The figure below shows a polygon 
, consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that 
and 
. 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]](http://latex.artofproblemsolving.com/8/c/0/8c0183579e57a39614f2c3be1824f286b6bf4b86.png)
Solution
While imagining the folding, 
goes on 
goes on 
and 
goes on 
So, 
and 
Also, 
becomes an edge parallel to 
so that means 
Since 
then 
So, the area of 
is 
If we let be the base, then the height is 
So, the volume is 
~aops-g5-gethsemanea2
Remark
After folding polygon 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]](http://latex.artofproblemsolving.com/3/1/2/312686e55a4760716ba0dafc8635f2217e326d69.png)
~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. 
