Area of Cutoff Triangles

Given $\Delta ABC$ and points $M,$ $N,$ $P$ on the sides $BC,$ $AC,$ and $AB,$ respectively. We are interested in three cutoff triangles, $NAP,$ $PBM,$ and $MCN.$

Prove that the area of at least one of triangles $NAP,$ $PBM,$ $MCN$ does not exceed $\frac{1}{4}[\Delta ABC].$

This is problem 15.40 from V. V. Prasolov, Problems in Geometry, v 2.

Solution

As usual $[F]$ denotes the area of shape $F.$ Set $S=[\Delta ABC],$ $S_A==[\Delta NAP],$ $S_B==[\Delta PBM],$ and $S_C==[\Delta MCN].$ Then by the sine formula for the area,

$\displaystyle\frac{S_A}{S}=\frac{AP\cdot AN}{AB\cdot AC},$ $\displaystyle\frac{S_B}{S}=\frac{BP\cdot BM}{AB\cdot BC},$ $\displaystyle\frac{S_C}{S}=\frac{CM\cdot CN}{BC\cdot AC}.$

It follows that

$\displaystyle\frac{S_A\cdot S_B\cdot S_C}{S^3}=\frac{AP\cdot BP}{AB^2}\cdot \frac{BM\cdot CM}{BC^2}\cdot\frac{CN\cdot AN}{AC^2}.$

Now apply $\displaystyle ab\le\frac{1}{4}(a+b)^2$ to the three fractions on the right:

$\displaystyle\frac{S_A}{S}\cdot \frac{S_B}{S}\cdot \frac{S_C}{S}\le\left(\frac{1}{4}\right)^3.$

Which says that not all fractions on the left may exceed $\displaystyle\frac{1}{4}.$

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