Difference between revisions of "2018 AMC 8 Problems/Problem 24"
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Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= \sqrt 3s</math> and <math>JI=\sqrt 2s</math>. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is <math>\frac{\sqrt 6s^2}{2}</math>. <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math> | Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= \sqrt 3s</math> and <math>JI=\sqrt 2s</math>. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is <math>\frac{\sqrt 6s^2}{2}</math>. <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math> | ||
| + | ==See Also== | ||
{{AMC8 box|year=2018|num-b=23|num-a=25}} | {{AMC8 box|year=2018|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:33, 21 November 2018
Problem 24
In the cube 
with opposite vertices 
and 
and 
are the midpoints of edges 
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)
Solution
Note that is a rhombus.
Let the side length of the cube be 
. By the Pythagorean theorem, 
and 
. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is 
. 
. Thus 
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. 
