Maths Olympiad Prep

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Problem 897

AMC 12 late, AIME early
Geometry Difficulty 4.7 Prove it Hong Kong competition problems · 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.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official 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

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