MathLabs

Problem 1

Let PP be a point inside a given triangle ABCABC, and let DD, EE, FF be the feet of the perpendiculars from PP to the lines BCBC, CACA, ABAB respectively. Find all positions of PP for which BCPD+CAPE+ABPF\dfrac{BC}{PD} + \dfrac{CA}{PE} + \dfrac{AB}{PF} is least.
Step 1 of 5: Split the triangle's area through PP
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In plain words

No matter where you stand inside the triangle, the three heights you drop to the sides always rebuild the same total area — that fixed budget is what Cauchy–Schwarz will exploit.

SABC=12(BC⋅PD+CA⋅PE+AB⋅PF)S_{ABC} = \tfrac12\big(BC\cdot PD + CA\cdot PE + AB\cdot PF\big)
Detailed analysis

Joining PP to the three vertices splits △ABC\triangle ABC into △PBC,△PCA,△PAB\triangle PBC,\triangle PCA,\triangle PAB, whose areas are 12BC⋅PD\tfrac12 BC\cdot PD, 12CA⋅PE\tfrac12 CA\cdot PE, 12AB⋅PF\tfrac12 AB\cdot PF since PD,PE,PFPD,PE,PF are the corresponding heights. Their sum is the fixed area SABCS_{ABC}, independent of where PP sits inside the triangle.