Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Inscribed circle and perimeter.

## Inscribed circle and perimeter – AIME I, 1999

The inscribed circle of triangle ABC is tangent to AB at P, and its radius is 21 given that AP=23 and PB=27 find the perimeter of the triangle

- is 107
- is 345
- is 840
- cannot be determined from the given information

**Key Concepts**

Inscribed circle

Perimeter

Triangle

## Check the Answer

But try the problem first…

Answer: is 345.

AIME I, 1999, Question 12

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

Q tangency pt on AC, R tangency pt on BC AP=AQ=23 BP=BR=27 CQ=CR=x and

Second Hint

\(s \times r =A\) and \(s=\frac{27 \times 2+23 \times 2+x \times 2}{2}=50+x\) and A=\(({(50+x)(x)(23)(27)})\) then from these equations 441(50+x)=621x then x=\(\frac{245}{2}\)

Final Step

perimeter 2s=2(50+\(\frac{245}{2}\))=345.

## 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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