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
Answer: is 345.
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.
Other useful links
- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA