Tetrahedron Problem | AIME I, 1992 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron.

Tetrahedron Problem - AIME I, 1992


Faces ABC and BCD of tetrahedron ABCD meet at an angle of 30,The area of face ABC=120, the area of face BCD is 80, BC=10. Find volume of tetrahedron.

  • is 107
  • is 320
  • is 840
  • cannot be determined from the given information

Key Concepts


Area

Volume

Tetrahedron

Check the Answer


Answer: is 320.

AIME I, 1992, Question 6

Coordinate Geometry by Loney

Try with Hints


First hint

Area BCD=80=\(\frac{1}{2} \times {10} \times {16}\),

where the perpendicular from D to BC has length 16.

Tetrahedron Problem

Second Hint

The perpendicular from D to ABC is 16sin30=8

[ since sin30=\(\frac{perpendicular}{hypotenuse}\) then height = perpendicular=hypotenuse \(\times\) sin30 ]

Final Step

or, Volume=\(\frac{1}{3} \times 8 \times 120\)=320.

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Largest Area of Triangle | AIME I, 1992 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Largest Area of Triangle.

Area of Triangle - AIME I, 1992


Triangle ABC has AB=9 and BC:AC=40:41, find the largest area that this triangle can have.

  • is 107
  • is 820
  • is 840
  • cannot be determined from the given information

Key Concepts


Ratio

Area

Triangle

Check the Answer


Answer: is 820.

AIME I, 1992, Question 13

Coordinate Geometry by Loney

Try with Hints


First hint

Let the three sides be 9, 40x, 41x

area = \(\frac{1}{4}\sqrt{(81^2-81x^2)(81x^2-1)} \leq \frac{1}{4}\frac{81^2-1}{2}\)

Second Hint

or, \(\frac{1}{4}\frac{81^2-1}{2}=\frac{1}{8}(81-1)(81+1)\)

Final Step

=(10)(82)

=820.

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