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
Inscribed circle
Perimeter
Triangle
Answer: is 345.
AIME I, 1999, Question 12
Geometry Vol I to IV by Hall and Stevens
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.
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