2018 AMC 8 Problems/Problem 24
Problem 24
In the cube 
with opposite vertices 
and 
and 
are the midpoints of segments 
and 
respectively. Let 
be the ratio of the area of the cross-section 
to the area of one of the faces of the cube. What is 
![[asy] size(6cm); pair A,B,C,D,EE,F,G,H,I,J; C = (0,0); B = (-1,1); D = (2,0.5); A = B+D; G = (0,2); F = B+G; H = G+D; EE = G+B+D; I = (D+H)/2; J = (B+F)/2; filldraw(C--I--EE--J--cycle,lightgray,black); draw(C--D--H--EE--F--B--cycle); draw(G--F--G--C--G--H); draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J); label("$A$",A,E); label("$B$",B,W); label("$C$",C,S); label("$D$",D,E); label("$E$",EE,N); label("$F$",F,W); label("$G$",G,N); label("$H$",H,E); label("$I$",I,E); label("$J$",J,W); [/asy]](http://latex.artofproblemsolving.com/a/1/d/a1d819cf8339fd4ae7e3d612b459da3b1e8cdc7b.png)
Note
In the 2008 AMC 10A, Question 21 was nearly identical to this question, except that in this question you have to look for the square of the area, not the actual area.
Video Solution
https://www.youtube.com/watch?v=04pV_rZw8bg - Happytwin
Video Solution
https://www.youtube.com/watch?v=ji9_6XNxyIc ~ MathEx
See Also
| 2018 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. 
