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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, 2011 based on Rectangles and sides.

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

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

$s \times r =A$ 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=$\frac{245}{2}$

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