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Geometry Difficulty 5.6 AIME, harder Prove it Argentina

In triangle ABCABC point PP divides side ABAB in ratio APPB=14\frac{AP}{PB} = \frac{1}{4}. The perpendicular bisector of segment PBPB intersects side BCBC at point QQ. If area(PQC)=425area(ABC)\text{area}(PQC) = \frac{4}{25}\text{area}(ABC) and AC=7AC = 7, find BCBC.

Solution

If area(ABC)=S\text{area}(ABC) = S then area(APC)=APABS=15S\text{area}(APC) = \frac{AP}{AB}S = \frac{1}{5}S. As area(PQC)=425S\text{area}(PQC) = \frac{4}{25}S, we have area(PQB)=S−15S−425S=1625S\text{area}(PQB) = S - \frac{1}{5}S - \frac{4}{25}S = \frac{16}{25}S. On the other hand
area(PBQ)=BQBCarea(PBC)=BQBC⋅BPBAarea(ABC)=BQBC⋅45S. \text{area}(PBQ) = \frac{BQ}{BC}\text{area}(PBC) = \frac{BQ}{BC} \cdot \frac{BP}{BA}\text{area}(ABC) = \frac{BQ}{BC} \cdot \frac{4}{5}S.
Hence 1625S=BQBC⋅45S\frac{16}{25}S = \frac{BQ}{BC} \cdot \frac{4}{5}S which implies BQBC=45\frac{BQ}{BC} = \frac{4}{5}. Because BPBA=45\frac{BP}{BA} = \frac{4}{5}, it follows that PQ∥ACPQ \parallel AC, so that triangles ABCABC and PBQPBQ are similar. But PQ=BQPQ = BQ as QQ belongs to the perpendicular bisector of PBPB. Therefore BC=AC=7BC = AC = 7.

Figure 1

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