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

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<asy>
 
<asy>
 
usepackage("mathptmx");
 
usepackage("mathptmx");
−
unitsize(1cm);
+
size(275);
−
defaultpen(linewidth(0.7)+fontsize(11));
+
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);
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dot("$D$",D,dir(0));
 
dot("$D$",D,dir(0));
 
dot("$E$",E,S);
 
dot("$E$",E,S);
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dot("$F$",F,1.5*S);
+
dot("$F$",F,1.5*dir(-100));
 
dot("$G$",G,S);
 
dot("$G$",G,S);
 
dot("$H$",H,W);
 
dot("$H$",H,W);
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</asy>
 
</asy>
  
−
<math>\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288\qquad</math>
+
<math>\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288</math>
  
 
==Solution==
 
==Solution==

Revision as of 02:07, 19 February 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 $24\cdot8=\boxed{\textbf{(C) }192}.$

Solution by aops-g5-gethsemanea2

Remark

It is easy to visualize the prism when $\triangle BJG$ is the base and $\overline{\rm AB}$ is the height.

~MathFun1000

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

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