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 4 of 7: Assume all three areas are too large
Previous stepNext step
In plain words

The target is a minimum statement, so its natural negation is that every one of the three candidates exceeds the threshold.

p(1−r)>14,q(1−p)>14,r(1−q)>14p(1-r)>\frac14,\quad q(1-p)>\frac14,\quad r(1-q)>\frac14
Detailed analysis

To prove that at least one area is at most 14[ABC]\frac14[ABC], assume for contradiction that all three are strictly larger. Using step 2 and step 3, this is exactly the displayed system of three strict inequalities.