In a triangle ABC, if certain areas are equal then P is its centroid

Question

Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$, have equal areas, prove that $P$ is the centroid of $\triangle ABC$.

I am trying to do with Ceva's theorem: $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$ and also with the result $\frac{AP}{PD}=\frac{AF}{FB}+\frac{AE}{EC}$ but having some difficulties. Please give any hint.

Answer 1

This is not a trivial geometry problem... Let $x=AF/FB, y=BD/DC, z=CE/EA$, then we have $xyz=1$. Furthermore, we can calculate that the area of $AFB$ is $\frac{x}{(1+z+yz)(1+x)}$. Similarly we can calculate two others, so, we have $$\frac{x}{(1+z+xz)(1+x)}=\frac{y}{(1+x+yx)(1+y)}=\frac{z}{(1+y+zy)(1+z)}$$ Notice that $(1+z+xz)=(xyz+z+xz)=z(xy+1+x)=z(xy+xyz+x)=zx(y+yz+1)$, so we have $$\frac{x}{(1+x)}=\frac{yz}{(1+y)}=\frac{z^2x}{(1+z)}$$ So we have $x(1+z)=z^2x(1+x)$ and $x(1+y)=yz(1+x)$. Substitute $y=1/xz$, we have $x(1+1/xz)=z/(xz)(1+x)$, or, $x(1+xz)=z(1+x)$. This yields a linear equation of $z$, that is, $z=\frac{x}{1+x-x^2}$. Substitute it in, we have $$(1+\frac{x}{1+x-x^2})=(\frac{x}{1+x-x^2})^2(1+x)$$ Or, $$(1+2x-x^2)(1+x-x^2)=x^2(1+x)$$ Expanding, we have $$x^4-4x^3-x^2+3x+1=0$$ Factorizing, we have $$(x-1)(x^3-3x^2-4x-1)=0$$ One solution is $x=1$, the other positive solution is located $x>4$, where $z=\frac x{1+x-x^2}<0$ (which means $P$ is outside), contradiction. Therefore, the only valid solution is $x=y=z=1$.

Answer 2

Let us give name $a$ for the area of triangles with common area and $x,y,z$ for the other areas. We have (using the following result: the areas of triangles sharing a same altitude with collinear bases are in the ratio of the lengths of these bases):

$$\begin{cases}\dfrac{EA}{EC}&=&\dfrac{x}{a}&=&\dfrac{a+z}{a+y}\\ \dfrac{DC}{DB}&=&\dfrac{y}{a}&=&\dfrac{a+x}{a+z}\\ \dfrac{FB}{FA}&=&\dfrac{z}{a}&=&\dfrac{a+y}{a+x}\end{cases}\tag{1}$$

By taking new variables

$$X=\dfrac{x}{a}, \ \ Y=\dfrac{y}{a}, \ \ Z=\dfrac{z}{a}$$

(1) becomes :

$$\begin{cases}X&=&\dfrac{1+Z}{1+Y}\\ Y&=&\dfrac{1+X}{1+Z}\\ Z&=&\dfrac{1+Y}{1+X}\end{cases}\tag{2}$$

1. The product of these 3 equations (2) gives

$$\underbrace{XYZ}_P=1 \tag{3}$$

2. Expanding the equations of (2) gives

$$\begin{cases} 1+Z&=&X+XY\\ 1+X&=&Y+YZ\\ 1+Y&=&Z+ZX \end{cases} \tag{4}$$

Adding equations (4) gives:

$$\underbrace{XY+YZ+ZX}_R=3 \tag{5}$$

Multiplying equations in (4), by $Z$, $X$ and $Y$ resp., we get:

$$\begin{cases} Z+Z^2&=&XZ+XYZ\\ X+X^2&=&YX+XYZ\\ Y+Y^2&=&ZY+XYZ \end{cases} \tag{6}$$

adding them and setting $S:=X+Y+Z$, we get:

$$S+(S^2-2R)=R+P$$

Taking into account (3) and (5), we get:

$$S+S^2-12=0, \tag{7}$$

a quadratic whose unique positive root is $S=3$.

Therefore, due to Vieta's relationships, $X,Y,Z$ are roots of the third degree equation in $T$ :

$$T^3-ST^2+RT-P=0 \ \iff \ T^3-3T^2+3T-1=0 \ \iff \ (T-1)^3=0\tag{8}$$

giving the triple root $X=Y=Z=1$. Otherwise said:

$$x=y=z=a\tag{9}$$

As a consequence, the areas of triangles $MAB, AMC, ABM$ are identical with common value $2a$.

Therefore, the barycentric coordinates of $M$ are $([MAB]/[ABC], [AMC]/[ABC], [ABM]/[ABC])=(1/3,1/3,1/3)$ characterizing the centroid of $ABC$.

Edit: I just discovered this similar question If three cevians are concurrent at a point and form triangles of equal area, the point is the centroid with an interesting answer by @almagest whose reasoning is partly the same as mine, with a final clever use of the Arithmetic/Geometric means inequality.

Answer 3

I will use barycentric coordinates, an excellent reference is

Barycentric Coordinates for the Impatient, Max Schindler, Evan Chen

and it is by chance compact, and easily accesible/readable. We will use the formula in Theorem 10 in it.

Let $P$ be $P=(x,y,z)$ in barycentric coordinates w.r.t. $\Delta ABC$, $x,y,z >0$, $x+y+z=1$. We may and do assume that the area of the given triangle is normed to $[ABC]=1$.

Then the area $[AFP]$ of $\Delta AFP$ is computed using $$ \begin{aligned} A &= (1,0,0)\ ,\\ P &= (x,y,z)\ , \\ F &= [x:y:0]=\left(\frac x{x+y},\frac y{x+y},0\right)\ ,\text{ so}\\ [AFP] &= \begin{vmatrix} 1 & 0 & 0\\ \frac x{x+y} &\frac y{x+y} & 0\\ x & y & z \end{vmatrix} = \frac 1{x+y} \begin{vmatrix} 1 & 0 & 0\\ x & y & 0\\ x & y & z \end{vmatrix} = \frac {yz}{x+y} = \frac {xyz}{x(x+y)} \ . \end{aligned} $$ The other two areas are obtained by cyclic permutation of the letters $x,y,z$ (taken in this order). Then the given equality of areas becomes: $$ \frac {xyz}{x(x+y)} = \frac {xyz}{y(y+z)} = \frac {xyz}{z(z+x)} \ . $$ So $x(x+y)=y(y+z)=z(z+x)$. Let $s$ be the common value of this expression, and let us obtain some further relation, here with all details... $$ \begin{aligned} s &= x(x+y) = x(1-z) = x - xz\ ,\\ s &= y(y+z) = y(1-x) = y - yx\ ,\\ s &= z(z+x) = z(1-y) = z - zy\ ,\\[2mm] 3s &= 1 - (xy+yz+zx)\ ,\\ s &= ys + zs + xs = y(x-xz)+z(y-yx) +x(z-zy)\\ &= (xy+yz+zx) -3xyz\ ,\\[2mm] s(1-3s) &= s(yz+zx+xy) =\sum yz(x-xz)= 3xyz - xyz = 2xyz\ ,\\ 3s(1-3s) &= 6xyz = 2(xy+yz+zx) - 2s = 2(1-3s)-2s\ , \end{aligned} $$ and we obtain an equation of degree two in $s$, which is $$ 0 = (2-3s)(1-3s) - 2s =2-11s + 9s^2=(1-s)(2-9s)\ . $$ Since $s=x(x+y)<1$, we only accept $s=2/9$. The simple elementary polynomials in $x,y,z$ are $$ \begin{aligned} e_1 &=x+y+z=1\ ,\\ e_2 &=xy+yz+zx=1-3s=\frac 13\ ,\\ e_3 &=xyz=\frac 12s(1-3s)=\frac 1{27}\ , \end{aligned} $$ so $x,y,z$ are (Vieta) the roots of the polynomial in $T$ $$ T^3 - T^2+\frac 13T -\frac 1{27}=\left(T-\frac 13\right)^3\ , $$ so the point $P$ is $P=(x,y,z)=\left(\frac 13,\frac 13,\frac 13\right)=[1:1:1]$, the centroid.