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

But try the problem first…

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.

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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