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Pyramid with Square base | AIME I, 1995 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Pyramid with Square base.

Pyramid with Squared base - AIME I, 1995


Pyramid OABCD has square base ABCD, congruent edges OA,OB,OC,OD and Angle AOB=45, Let \(\theta\) be the measure of dihedral angle formed by faces OAB and OBC, given that cos\(\theta\)=m+\(\sqrt{n}\), find m+n.

  • is 107
  • is 5
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisibility

Algebra

Check the Answer


Answer: is 5.

AIME I, 1995, Question 12

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

Let \(\theta\) be angle formed by two perpendiculars drawn to BO one from plane ABC and one from plane OBC.

Let AP=1 \(\Delta\) APO is a right angled isosceles triangle, OP=AP=1.

Pyramid with square base

Second Hint

then OB=OA=\(\sqrt{2}\), AB=\(\sqrt{4-2\sqrt{2}}\), AC=\(\sqrt{8-4\sqrt{2}}\)

Final Step

taking cosine law

\(AC^{2}=AP^{2}+PC^{2}-2(AP)(PC)cos\theta\)

or, 8-4\(\sqrt{2}\)=1+1-\(2cos\theta\) or, cos\(\theta\)=-3+\(\sqrt{8}\)

or, m+n=8-3=5.

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