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 3 of 7: Express the other two corner areas
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In plain words

The cyclic structure of the three corners produces the same formula three times, with the variables shifted from (p,r)(p,r) to (q,p)(q,p) to (r,q)(r,q).

[BKM][ABC]=q(1−p),[CLK][ABC]=r(1−q)\frac{[BKM]}{[ABC]}=q(1-p),\quad \frac{[CLK]}{[ABC]}=r(1-q)
Detailed analysis

The same shared-angle argument at BB and CC gives [BKM][ABC]=BKBC⋅BMBA=q(1−p)\frac{[BKM]}{[ABC]}=\frac{BK}{BC}\cdot\frac{BM}{BA}=q(1-p) and [CLK][ABC]=CLCA⋅CKCB=r(1−q)\frac{[CLK]}{[ABC]}=\frac{CL}{CA}\cdot\frac{CK}{CB}=r(1-q).