SOLUTION: The figure shows a 2×2×2 cube ABCDEFGH, as well as midpoints I and J of its edges DH and BF. It so happens that C,I,E, and J all lie in a plane. Can you justify this statement?

I did that problem a week or so ago. 
Question 1163283



The figure ABOVE shows a 2 × 2 × 2 cube ABCDEFGH, as well as midpoints I and
J of its edges DH and BF. It so happens that C , I , E , and J all lie in a
plane. Can you justify this statement?
Yes I can. A quadrilateral is coplanar if and only if the sum of its
interior angles is 360°.  The plane of square ABFE is perpendicular to the
plane of BCGF. If two planes are perpendicular, then any angle whose sides
are in the two planes are 90°, so ∠CJE=90°  Thus EJ ∥ CJ and
similarly the other 4 interior angles of quadrilateral CIEJ also 90°. Thus
quadrilateral CIEF is also a rectangle.  Thus the sum of the interior angles is 
4∙90° = 360°.
What kind of figure is quadrilateral CIEJ,
It is not only a rectangle but also a square because right triangles ΔIHE,
ΔBJC, ΔJFE, ΔDIC are all congruent and their longer legs are its sides.
and what is its area?
Congruent right triangles ΔIHE, ΔBJC, ΔJFE, ΔDIC all have hypotenuses 2 and
shorter legs 1, so by the Pythagorean theorem, each side of square CIEJ is
√5.  Therefore the area of square CIEJ is 5.
Is it possible to obtain a polygon with a larger area by slicing the cube

with a different plane? If so, show how to do it. If not, explain why it is

not possible.
Yes, it is possible. CDEF and BGHA are rectangles with length 2√2 and width
2. So their areas are 4√2 which is approximately 5.656854249 > 5. 



Edwin