2018 AMC 8

Complete problem set with solutions and individual problem pages

Problem 24 Hard

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?

  • A.

    \frac54

  • B.

    \frac43

  • C.

    \frac32

  • D.

    \frac{25}{16}

  • E.

    \frac94

Answer:C

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

If the edges of the cube have same lengths a, A is the origin, \vec{AD } is the positive x direction, \vec{AE} is the positive y direction, and \vec{AB } is the positive z direction. Therefore, we have E(0,a,0), I(a,\frac{a}{2},0), C(a,0,a), and J(0,\frac{a}{2},a). Hence, we can figure out that:

|EC|=\sqrt{a^2 + a^2 + a^2}=\sqrt{3}a

|IJ|=\sqrt{a^2 + 0 + a^2}=\sqrt{2}a

Note that EJCI is a rhombus, so R=\frac{\sqrt{6}a^2}{2a^2}=\frac{\sqrt{6}}{2}. Finally, we can see that the answer is R^2=\frac{6}{4}=\boxed{\textbf{(C) } \frac{3}{2}}

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