Try this beautiful problem from Geometry: Ratio of the area of cube's cross section . You may use sequential hints to solve the problem.
In the cube ABCDEFGH with opposite vertices C and E ,J and I are the mid points of segments FB and 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 $R^2$ ?
Geometry
Area
Pythagorean theorem
Answer:$\frac{3}{2}$
AMC-8(2018) Problem 24
Pre College Mathematics
EJCI is a rhombus by symmetry
Can you now finish the problem ..........
Area of rhombus is half product of its diagonals....
can you finish the problem........
Let Side length of a cube be x.
then by the pythagorean theorem$ EC=X \sqrt {3}$
$JI =X \sqrt {2}$
Now the area of the rhombus is half product of its diagonals
therefore the area of the cross section is $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$
This shows that $R= \frac{\sqrt6}{2}$
i.e$ R^2=\frac{3}{2}$
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