Maths Olympiad Prep

Library / /7 of 48

Geometry Difficulty 4.7 AIME Prove it Hong Kong

Given triangle ABCABC, let DD be an inner point of the segment BCBC. Let PP and QQ be distinct inner points of the segment ADAD. Let K=BPACK = BP \cap AC, L=CPABL = CP \cap AB, E=BQACE = BQ \cap AC, F=CQABF = CQ \cap AB. Given that KLEFKL \parallel EF, find all possible values of the ratio BD:DCBD : DC.

Solution

The only possible value is 11.
We use projective geometry. Consider the following projection.
AB(A,L,F,B){C}AD(A,P,Q,D){B}AC(A,K,E,C). AB(A, L, F, B) \xrightarrow{\{C\}} AD(A, P, Q, D) \xrightarrow{\{B\}} AC(A, K, E, C).
This shows (A,L,F,B)(A, L, F, B) and (A,K,E,C)(A, K, E, C) are perspective, and so LKLK, FEFE, BCBC are concurrent. Since KLEFKL \parallel EF, this shows KLEFCBKL \parallel EF \parallel CB.
Now, since ADAD, BKBK, CLCL are concurrent, KLCBKL \cap CB is the harmonic conjugate of DD with respect to BCBC. As KLCBKL \cap CB is a point at infinity, DD must be the midpoint of BCBC.

Figure 1

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: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.