Categories

# Incentre and Triangle | AIME I, 2001 | Question 7

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

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

But try the problem first…

Source

AIME I, 2001, Question 7

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

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.

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.