Area Inequalities in Triangle

Given $\Delta ABC$ and points $M,$ $N,$ $P$ on the sides $BC,$ $AC,$ and $AB,$ respectively. There emerge four triangles, $NAP,$ $PBM,$ $MCN,$ and $MNP$ that combine into $\Delta ABC.$

The areas of the four triangles add up to that of $\Delta ABC$ such that one would expect that those areas may be somehow compared to the one fourth of the area of $\Delta ABC.$ Indeed, there are several instructive problems that give partial answer to the question of comparison, depending on the manner of construction of points $M,$ $N,$ $P.$ As usual, $[F]$ stands for the area of shape $F.$

Problem 1

Let, for a positive $k,$ $AP=k\cdot CN$ and $BP=k\cdot AN.$

Area Inequalities in Triangle, problem 1

Show that $[\Delta NAP]\le \frac{1}{4}[\Delta ABC].$

By the sine formula for the area,

$\displaystyle \frac{[\Delta NAP]}{[\Delta ABC]}=\frac{AN\cdot AP}{AB\cdot AC}.$

But $AP=k\cdot CN$ and $BP=k\cdot AN.$ It follows, by the AM-GM inequality, that,

$\displaystyle\begin{align} \frac{AN\cdot AP}{AB\cdot AC} &= \frac{AN\cdot AP}{(AP+k\cdot AN)(AN+\frac{1}{k}AP)}\\ &=\frac{(k\cdot AN)\cdot AP}{(AP+k\cdot AN)(AP+k\cdot AN)}\\ &=\left(\frac{1}{2}\right)^2. \end{align}$

(This is problem 15.39 (with k=1) from V. V. Prasolov, Problems in Geometry, v 2.)

Problem 2

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

The solution to this problem can be found on a separate page.

Problem 3

Let $AM,$ $BN,$ $CP$ be concurrent cevians.

Area Inequalities in Triangle, problem 3

Prove that $[\Delta MNP]\le \frac{1}{4}[\Delta ABC].$

The solution to this problem can be found on a separate page.

To remind, $\Delta MNP$ is known as a cevian triangle of the point of the intersection of the cevians. Due to Ceva's theorem, the premise of the problem is equivalent to the condition:

$AP\cdot BM\cdot CN = PB\cdot MC\cdot NA.$

Problem 4

Let $M,$ $N,$ $P$ be the pedal points of point $D.$

Area Inequalities in Triangle, problem 4

Prove that $[\Delta MNP]\le \frac{1}{4}[\Delta ABC].$

To remind, in this situation, $\Delta MNP$ is known as a pedal triangle of point $D$ with respect to $\Delta ABC.$

The area of the pedal triangle is known to be

$\displaystyle [\Delta MNP]=\frac{1}{4}(R^2-OD^2)\frac{[\Delta ABC]}{R^2},$

where $O$ and $R$ are the circumcenter and circumradius of $\Delta ABC.$ It is obvious that $\displaystyle \frac{R^2-OD^2}{R^2}\le 1,$ thus solving Problem 4. The identity is achieved only for the pedal triangle of the circumcenter.

Problem 5

Assume points $M,$ $N,$ $P$ have the property

$AP + BM + CN = PB + MC + NA.$

Then $[\Delta MNP]\le \frac{1}{4}[\Delta ABC].$

The solution to this problem can be found on a separate page.

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