Maths Olympiad Prep

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Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Hong Kong

Let ADAD, BEBE and CFCF be respectively the altitudes of ABC\triangle ABC. Let PP, QQ and RR be points on the lines BCBC, CACA and ABAB respectively such that APAP is perpendicular to EFEF, BQBQ is perpendicular to FDFD and CRCR is perpendicular to DEDE. Show that APAP, BQBQ and CRCR are concurrent.

Solution

By a corollary of Carnot's theorem, the perpendicular lines drawn from AA, BB, CC to EFEF, FDFD, DEDE are concurrent if and only if the perpendicular lines drawn from DD, EE, FF to BCBC, CACA, ABAB are concurrent. The latter is clearly true since the perpendicular lines are just ADAD, BEBE, CFCF, which are concurrent at the orthocentre of ABC\triangle ABC. This proves APAP, BQBQ, CRCR are concurrent.

Figure 1

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