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

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.

## Other useful links

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