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2018 AMC 8 Problems/Problem 24

Revision as of 18:36, 6 November 2020 by Franzliszt (talk | contribs) (→‎Solution)

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 1

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 $R^2 = \boxed{\textbf{(C) } \frac{3}{2}}$

Solution 2

We can solve this with 3D Cartesian coordinates. Assume WLOG that the sides of the square are of length $2$. Let $C$ be the origin and let $\vec{CD}$ be the positive $x$ direction, $\vec{CG}$ be the positive $y$ direction, and $\vec{CB}$ be the positive $z$ direction. We find that $I=(2,1,0),J=(0,1,2),E=(2,2,2)$.

Notice that $CI=CJ=EI=EJ=\sqrt{5}$ so $EJCI$ is a rhombus. Furthermore, by the distance formula, $IJ=\sqrt8$.

By the Law of Cosines on $\triangle JCI$ we have $\cos \angle JCI=\frac{\sqrt5^2+ \sqrt5^2-\sqrt8^2}{2\cdot \sqrt5\cdot \sqrt5}=\frac 15$. By the Law of Cosines on $\triangle JEI$ we have $\cos \angle JEI=\frac{\sqrt5^2+ \sqrt5^2-\sqrt8^2}{2\cdot \sqrt5\cdot \sqrt5}=\frac 15$.

Bretschneider's formula states given a quadrilateral $ABCD$ with sides $a,b,c,d$ then \[[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)}\]where $s=\frac{a+b+c+d}2$. Using this formula, we find that\[[EJCI]=\sqrt{(2\sqrt5-\sqrt5)(2\sqrt5-\sqrt5)(2\sqrt5-\sqrt5)(2\sqrt5-\sqrt5)-\sqrt 5^4(\tfrac 15)^2}=2\sqrt{6}.\]

Using Bretschneider's formula again, we can find that $[ABCD]=\sqrt{(4-2)(4-2)(4-2)(4-2)-2^4\cdot \cos \left(\frac{90^\circ+90^\circ}2\right)}=\sqrt{16}=4$.

The answer is thus $\left(\frac{2\sqrt6}{4}\right)^2=\frac 32$ so we circle answer choice $C$.

~franzliszt

Note

Problem 21 of the 2008 AMC 10A 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

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