# 2018 AMC 8 Problems/Problem 24

## Problem 24

In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of poop $\overline{FB}$ 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, $EC= \sqrt 3s$ and $JI=\sqrt 2s$. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is $\frac{\sqrt 6s^2}{2}$. This gives $R = \frac{\sqrt 6}2$. Thus $R^2 = \boxed{\textbf{(C) } \frac{3}{2}}$

## Note

In the 2008 AMC 10A, Question 21 (https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_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 byProblem 23 Followed byProblem 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.

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