Maths Olympiad Prep

Library / /109 of 353

Geometry Difficulty 4.8 AIME Find the answer

The three points A, B, C form a triangle. AB=4, BC=5, AC=6. Let the angle bisector of ∠\angle A intersect side BC at D. Let the foot of the perpendicular from B to the angle bisector of ∠\angle A be E. Let the line through E parallel to AC meet BC at F. Compute DF.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since AD bisects ∠\angle A, by the angle bisector theorem ABBD=ACCD,\frac{AB}{BD}=\frac{AC}{CD}, so BD=2 and CD=3. Extend BE to hit AC at X. Since AE is the perpendicular bisector of BX, AX=4. Since B, E, X are collinear, applying Menelaus' Theorem to the triangle ADC, we have AEED⋅DBBC⋅CXXA=1.\frac{AE}{ED} \cdot \frac{DB}{BC} \cdot \frac{CX}{XA}=1. This implies that AEED=5,\frac{AE}{ED}=5, and since EF ∥\parallel AC, DFDC=DEDA,soDF=DC6=12.\frac{DF}{DC}=\frac{DE}{DA}, so DF=\frac{DC}{6}=\frac{1}{2}.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.