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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.