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# Inscribed circle and perimeter | AIME I, 1999 | Question 12

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

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.

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