Difference between revisions of "2022 AMC 8 Problems/Problem 24"
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<asy> | <asy> | ||
usepackage("mathptmx"); | usepackage("mathptmx"); | ||
− | + | size(275); | |
− | defaultpen(linewidth(0. | + | defaultpen(linewidth(0.8)); |
real r = 2, s = 2.5, theta = 14; | 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 G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta); | ||
Line 19: | Line 19: | ||
dot("$D$",D,dir(0)); | dot("$D$",D,dir(0)); | ||
dot("$E$",E,S); | dot("$E$",E,S); | ||
− | dot("$F$",F,1.5* | + | dot("$F$",F,1.5*dir(-100)); |
dot("$G$",G,S); | dot("$G$",G,S); | ||
dot("$H$",H,W); | dot("$H$",H,W); | ||
Line 26: | Line 26: | ||
</asy> | </asy> | ||
− | <math>\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288 | + | <math>\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288</math> |
==Solution== | ==Solution== | ||
− | + | 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>BJ=CI=8</math> and <math>FG=BC=8.</math> Also, <math>\overline{HJ}</math> becomes an edge parallel to <math>\overline{FG},</math> so that means <math>HJ=8.</math> | |
− | Since <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: | ||
+ | <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> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/oUEa7AjMF2A?si=IKVkcHgWtsS8Fmbs&t=4943 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution(🚀Under 2 min🚀 With color-coded folding explanation ✨)== | ||
+ | https://youtu.be/FhQROS_83iw | ||
+ | |||
+ | <i>~Education, the Study of Everything</i> | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/MOcX5BFbcwU?t=380 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=2uoBPp4Kxck | ||
+ | |||
+ | ~Mathematical Dexterity | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Ij9pAy6tQSg?t=2432 | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=tQecx6F1O8k | ||
+ | |||
+ | ~David | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/0orAAUaLIO0?t=469 | ||
+ | |||
+ | ~STEMbreezy | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/YJceP3zGSEU | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/_yj11gFIdw4 | ||
+ | |||
+ | Please like and subscribe! | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=23|num-a=25}} | {{AMC8 box|year=2022|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:03, 23 November 2023
Contents
[hide]- 1 Problem
- 2 Solution
- 3 Remark
- 4 Video Solution by Math-X (First understand the problem!!!)
- 5 Video Solution(🚀Under 2 min🚀 With color-coded folding explanation ✨)
- 6 Video Solution by OmegaLearn
- 7 Video Solution
- 8 Video Solution
- 9 Video Solution
- 10 Video Solution
- 11 Video Solution
- 12 Video Solution
- 13 See Also
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?
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: ~MRENTHUSIASM
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=IKVkcHgWtsS8Fmbs&t=4943
~Math-X
Video Solution(🚀Under 2 min🚀 With color-coded folding explanation ✨)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/MOcX5BFbcwU?t=380
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=2uoBPp4Kxck
~Mathematical Dexterity
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=2432
~Interstigation
Video Solution
https://www.youtube.com/watch?v=tQecx6F1O8k
~David
Video Solution
https://youtu.be/0orAAUaLIO0?t=469
~STEMbreezy
Video Solution
~savannahsolver
Video Solution
Please like and subscribe!
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.