Problem 6

Let

M

K

, and

L

be points chosen in the interiors of sides

AB

BC

, and

CA

, respectively, of triangle

ABC

. Prove that the area of at least one of the three triangles

\triangle AML

\triangle KBM

, and

\triangle LCK

is less than or equal to one-fourth of the area of triangle

ABC

Step 1 of 7: Parameterize the three side points

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In plain words

Ratios remove the shape and size of $ABC$ : only the relative positions of $M,K,L$ matter for the three area fractions.

p=\frac{AM}{AB},\quad q=\frac{BK}{BC},\quad r=\frac{CL}{CA},\quad 0<p,q,r<1

Detailed analysis

Set $p=\frac{AM}{AB}$ , $q=\frac{BK}{BC}$ , and $r=\frac{CL}{CA}$ . Since the points are interior points, $0<p,q,r<1$ , and the complementary ratios are $\frac{BM}{AB}=1-p$ , $\frac{CK}{BC}=1-q$ , and $\frac{AL}{CA}=1-r$ .