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## 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

Area

Volume

Tetrahedron

## Check the Answer

Answer: is 320.

AIME I, 1992, Question 6

Coordinate Geometry by Loney

## Try with Hints

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

where the perpendicular from D to BC has length 16.

The perpendicular from D to ABC is 16sin30=8

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

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

Ratio

Area

Triangle

## Check the Answer

Answer: is 820.

AIME I, 1992, Question 13

Coordinate Geometry by Loney

## Try with Hints

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

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

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

=(10)(82)

=820.