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

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

Inscribed circle

Perimeter

Triangle

But try the problem first...

Answer: is 345.

Source

Suggested Reading

AIME I, 1999, Question 12

Geometry Vol I to IV by Hall and Stevens

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.

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

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