Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2001 based on Incentre and triangle.
Incentre and Triangle – AIME I, 2001
Triangle ABC has AB=21 AC=22 BC=20 Points D and E are located on AB and AC such that DE parallel to BC and contains the centre of the inscribed circle of triangle ABC then \(DE=\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n
- is 107
- is 923
- is 840
- cannot be determined from the given information
Key Concepts
Incentre
Triangles
Algebra
Check the Answer
Answer: is 923.
AIME I, 2001, Question 7
Geometry Vol I to IV by Hall and Stevens
Try with Hints
F is incentre, BF and CF are angular bisectors of angle ABC and angle ACB DE drawn parallel to BC then angle BFD=angle FBC= angle FBD
triangle BDF is isosceles then same way triangle CEF is isosceles then perimeter triangle ADE=AD+DE+AE=AB+AC=43 perimeter triangle ABC=63 and triangle ABC similar to triangle ADE
DE=\(BC \times \frac{43}{63}\)=\(20 \times \frac{43}{63}\)=\(\frac{860}{63}\) then m+n=860+63=923.
Other useful links
- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA