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Difference between revisions of "2018 AMC 8 Problems/Problem 24"

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==Problem 24==
 
==Problem 24==
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In the cube <math>ABCDEFGH</math> with opposite vertices <math>C</math> and <math>E,</math> <math>J</math> and <math>I</math> are the midpoints of vertices <math>\overline{FB}</math> and <math>\overline{HD},</math> respectively. Let <math>R</math> be the ratio of the area of the cross-section <math>EJCI</math> to the area of one of the faces of the cube. What is <math>R^2?</math>
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In the cube <math>ABCDEFGH</math> with opposite vertices <math>C</math> and <math>E,</math> <math>J</math> and <math>I</math> are the midpoints of segments <math>\overline{FB}</math> and <math>\overline{HD},</math> respectively. Let <math>R</math> be the ratio of the area of the cross-section <math>EJCI</math> to the area of one of the faces of the cube. What is <math>R^2?</math>
  
 
<asy>
 
<asy>

Revision as of 08:25, 11 November 2019

Problem 24

In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of segments $\overline{FB}$ and $\overline{HD},$ respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2?$

[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]

$\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}$

Solution

Note that $EJCI$ is a rhombus by symmetry. Let the side length of the cube be $s$. By the Pythagorean theorem, $EC= s\sqrt 3$ and $JI= s\sqrt 2$. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is $\frac{s^2\sqrt 6}{2}$. This gives $R = \frac{\sqrt 6}2$. Thus $R^2 = \boxed{\textbf{(C) } \frac{3}{2}}$

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.

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

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