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

(Problem 24)
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==Problem 24==
 
==Problem 24==
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 poop <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>
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Note that <math>EJCI</math> is a rhombus by symmetry.
 
Note that <math>EJCI</math> is a rhombus by symmetry.
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 its diagonals, the area of the cross section is <math>\frac{\sqrt 6s^2}{2}</math>. This gives <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math>
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Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= s\sqrt 3</math> and <math>JI= s\sqrt 2</math>. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is <math>\frac{s^2\sqrt 6}{2}</math>. This gives <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math>
  
 
==Note==
 
==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.
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In the 2008 AMC 10A,[[2008 AMC 10A Problems/Problem 21| 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.
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==Video Solution==
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https://www.youtube.com/watch?v=04pV_rZw8bg
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- Happytwin
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==Video Solution==
 +
https://www.youtube.com/watch?v=ji9_6XNxyIc ~ MathEx
  
 
==See Also==
 
==See Also==

Revision as of 21:17, 2 August 2020

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.

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AJHSME/AMC 8 Problems and Solutions

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