INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More

Content

[hide]

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2001 based on Incentre and triangle.

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

Incentre

Triangles

Algebra

But try the problem first...

Answer: is 923.

Source

Suggested Reading

AIME I, 2001, Question 7

Geometry Vol I to IV by Hall and Stevens

First hint

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

Second Hint

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

Final Step

DE=\(BC \times \frac{43}{63}\)=\(20 \times \frac{43}{63}\)=\(\frac{860}{63}\) then m+n=860+63=923.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
Google