MathLabs

Problem 6

Let MM, KK, and LL be points chosen in the interiors of sides ABAB, BCBC, and CACA, respectively, of triangle ABCABC. Prove that the area of at least one of the three triangles △AML\triangle AML, △KBM\triangle KBM, and △LCK\triangle LCK is less than or equal to one-fourth of the area of triangle ABCABC.
Step 7 of 7: Conclude the area inequality
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In plain words

A contradiction to the negation of the claim proves that at least one of the three areas meets the quarter-area bound.

min⁡{[AML],[BKM],[LCK]}≤14[ABC]\min\{[AML],[BKM],[LCK]\}\le\frac14[ABC]
Detailed analysis

The assumption that all three corner areas exceed 14[ABC]\frac14[ABC] is impossible. Therefore at least one of [AML][AML], [BKM][BKM], and [LCK][LCK] is at most 14[ABC]\frac14[ABC], as required.