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AMC-8 Geometry Math Olympiad USA Math Olympiad

Area of cube’s cross section |Ratio | AMC 8, 2018 – Problem 24

Try this beautiful problem from Geometry: Ratio of the area of cube’s cross section . You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry: Ratio of the area of cube’s cross section . You may use sequential hints to solve the problem.

Area of cube’s cross section – AMC-8, 2018 – Problem 24


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 $R^2$ ?

area of cube's cross section

  • $\frac{5}{4}$
  • $\frac{3}{2}$
  • $\frac{4}{3}$

Key Concepts


Geometry

Area

Pythagorean theorem

Check the Answer


But try the problem first…

Answer:$\frac{3}{2}$

Source
Suggested Reading

AMC-8(2018) Problem 24

Pre College Mathematics

Try with Hints


First hint

EJCI is a rhombus by symmetry

Can you now finish the problem ……….

Second Hint

Area of rhombus is half product of its diagonals….

can you finish the problem……..

Final Step

area of cube's cross section

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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