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 2 of 7: Express the area of △AML\triangle AML
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In plain words

Sharing an angle means the sine factor cancels, so a two-dimensional area ratio becomes a product of one-dimensional side ratios.

[AML][ABC]=p(1−r)\frac{[AML]}{[ABC]}=p(1-r)
Detailed analysis

Triangles AMLAML and ABCABC share the angle at AA, so the area ratio is the product of the two adjacent side ratios: [AML][ABC]=AMAB⋅ALAC=p(1−r)\frac{[AML]}{[ABC]}=\frac{AM}{AB}\cdot\frac{AL}{AC}=p(1-r).