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


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

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


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

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

Sometimes it is more convenient to have this expression in different forms. If $\displaystyle x=\frac{\alpha}{\beta},\,$ $\displaystyle y=\frac{\beta}{\gamma},\,$ $\displaystyle z=\frac{\gamma}{\alpha},$ Then

$\displaystyle [\Delta MNP]=\frac{2}{(x+1)(y+1)(z+1)}[\Delta ABC].$

On the other hand, if we find, $x,y,z\,$ such that $\alpha:\beta=(1-x):x,\,$ $\beta:\gamma=(1-y):y\,$ and $\gamma:\alpha=(1-z):z\,$ then

$\displaystyle [\Delta MNP]=2xyz[\Delta ABC].$

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

  1. 3D Quadrilateral - a Coffin Problem
  2. Barycentric Coordinates
  3. Barycentric Coordinates: a Tool
  4. Barycentric Coordinates and Geometric Probability
  5. Ceva's Theorem
  6. Determinants, Area, and Barycentric Coordinates
  7. Maxwell Theorem via the Center of Gravity
  8. Bimedians in a Quadrilateral
  9. Simultaneous Generalization of the Theorems of Ceva and Menelaus
  10. Three glasses puzzle
  11. Van Obel Theorem and Barycentric Coordinates
  12. 1961 IMO, Problem 4. An exercise in barycentric coordinates
  13. Centroids in Polygon
  14. Center of Gravity and Motion of Material Points
  15. Isotomic Reciprocity
  16. An Affine Property of Barycenter
  17. Problem in Direct Similarity
  18. Circles in Barycentric Coordinates
  19. Barycenter of Cevian Triangle
  20. Concurrent Chords in a Circle, Equally Inclined

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