Area of Cevian Triangle

Let $AM,$ $BN,$ $CP$ be concurrent cevians in $\Delta ABC.$ Assume that point $D$ of concurrence has the barycentric coordinates $(\alpha :\beta :\gamma),$ $\alpha +\beta +\gamma=1.$

Area of Cevian Triangle, problem 3

Then

$\displaystyle[\Delta MNP] = \frac{2\alpha\beta\gamma}{(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )}[\Delta ABC],$

where $[F]$ denotes the area of shape $F.$

Proof

Elsewhere we derived the formula for the area of $\Delta K_1K_2K_3,$ where $K_i =(u_i:v_i:w_i),$ $u_i+v_i+w_i=1,$ $i=1,2,3:$

$[\Delta K_1K_2K_3]=\left| \begin{array} \;u_1 & v_1 & w_1\\ u_2 & v_2 & w_2\\ u_3 & v_3 & w_3 \end{array} \right|[\Delta ABC].$

For $D=(\alpha :\beta :\gamma),$ $M=(0:\beta :\gamma),$ $N=(\alpha :0:\gamma),$ $P=(\alpha :\beta :0)$ which after the normalization appear as

$\displaystyle M=\frac{1}{\beta +\gamma}(0:\beta :\gamma),$ $\displaystyle N=\frac{1}{\alpha +\gamma}(\alpha :0:\gamma),$ $\displaystyle P=\frac{1}{\alpha +\beta}(\alpha :\beta :0)$

the formula gives

$\displaystyle\begin{align} [\Delta MNP] &=\frac{1}{(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )}\left| \begin{array} \;0 & \beta & \gamma\\ \alpha & 0 & \gamma\\ \alpha & \beta & 0 \end{array} \right|[\Delta ABC]\\ &=\frac{2\alpha\beta\gamma}{(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )}[\Delta ABC]. \end{align}$

It is now easy to estimate the area of the cevian triangle:

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

This follows from the inequality for $a,b,c\gt 0:$

$(a+b)(b+c)(c+a)\ge 8abc,$

with equality only when $a=b=c.$ Indeed, by the AM-GM inequality

$\displaystyle\frac{a+b}{2}\frac{b+c}{2}\frac{c+a}{2}\ge \sqrt{ab}\sqrt{bc}\sqrt{ac}=abc.$

It follows that the only time when the area of a cevian triangle equals $\displaystyle\frac{1}{4}[\Delta ABC]$ is when $D=(1:1:1),\;$ i.e., when $D$ is the centroid of $\Delta ABC.$

Barycenter and Barycentric Coordinates

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