Triangle center dividing it in three quadrilaterals of the same area

Question

Inspired by this problem, I'd like to ask

Given a triangle $\triangle ABC$, what is the point $X$ on the plane so that $$[AEXF] = [BFXD] = [CDXE]$$ where $D, E, F$ are the orthogonal projections of $X$ onto $BC, CA, AB$, respectively.

By "what is the point" I mean to ask for its name (because knowing a point's name gives you power over it). I mean, triangle centers have been studied exhaustively so I'd be very surprised if such a simple setup hasn't come up in the past. It'd be also interesting to know how to construct it and its coordinates in terms of $A, B, C$.

For starting some work, let's name the relevant lengths as in the image above. $$\begin{cases} \overline{XD} = x,\qquad\overline{BC} = a,\qquad\overline{CA} = b,\qquad\overline{AB} = c\\ \overline{XE} = y,\qquad\overline{CD} = i,\qquad\overline{AE} = k,\qquad\overline{BF} = m,\\ \overline{XF} = z,\qquad\overline{BD} = j,\qquad\overline{CE} = l, \qquad\overline{AF} = n\end{cases}$$

There are several ways to write the areas of the quadrilaterals, each one giving us a set of equivalent equations.

---

$$\begin{cases}~ [AEXF] &= \dfrac12(ky+nz)\\ [BFXD] &= \dfrac12(mz+jx)\\ [CDXE] &= \dfrac12(ix+ly) \end{cases}$$ $$ \implies ky+nz = mz+jx = ix+ly $$

---

$$\begin{cases}~ [AEXF] &= \dfrac12\sin A(kn+yz)\\ [BFXD] &= \dfrac12\sin B(mj+zx)\\ [CDXE] &= \dfrac12\sin C(il+xy) \end{cases}$$ $$\begin{aligned} &\implies \sin A(kn+yz) = \sin B(mj+zx) = \sin C(il+xy)\\ &\implies a(kn+yz) = b(mj+zx) = c(il+xy) \end{aligned}$$ The second implication above is a consequence of the law of sines.

Answer 1

1. Define the vectors $v_1 = AB, v_2 = BC, v_3 = CA$

2. Normalize these vectors into $u_1, u_2, u_3$ respectively.

3. Find the perpendicular vectors $w_1, w_2, w_3$, respectively, by rotating $u_i$ $90^\circ$ clockwise , $ i = 1, 2, 3 $.

4. Let the point we're looking for be $P = (x_1, y_1) $

5. Referring to the figure in the OP, it follows that

$n = u_1 \cdot (P - A) $

$z = w_1 \cdot (P - A) $

$m = - u_1 \cdot (P - B)$

$j = u_2 \cdot (P - B) $

$x = w_2 \cdot (P - B) $

$i = - u_2 \cdot (P - C) $

$ l = u_3 \cdot (P - C) $

$ y = w_3 \cdot (P - C) $

$ k = - u_3 \cdot (P - A) $

These are all the distances we need

6. Now, we have the following area expressions,

$[ABC] = \dfrac{1}{2} ( u_3 \cdot (A - C) )( w_3 \cdot (B - A) ) $

Using the transpose operator, this can be written as

$ [ABC] = \dfrac{1}{2} (A - C)^T u_3 w_3^T (B - A) $

7. Similarly, and as listed in the OP, we find that

$[AEPF] = \dfrac{1}{2} ( k y + n z ) = \dfrac{1}{3} [ABC]$

Therefore, the first equation we have is,

$ \dfrac{1}{2} ( - (P - A)^T u_3 w_3^T (P - C) + (P - A) u_1 w_1^T (P - A) ) = \dfrac{1}{3} [ABC] $

8. The second equation we have is, similarly,

$[BFPD] = \dfrac{1}{2} ( mz + jx) = \dfrac{1}{3}[ABC]$

Therefore,

$ \dfrac{1}{2} ( - (P - B)^T u_1 w_1 (P - A) + (P - B) u_2 w_2^T (P - B) ) = \dfrac{1}{3} [ABC] $

Both area expressions are quadratic equations in $x_1, y_1$, the coordinates of $P$.

9. Now all we need is a two-variable quadratic system solver, and this I have.

10. Example I took $A = (3, 2) , B = (8,12), C = (17, 5) $. The solver produced four real solutions but only one of them is valid with $P$ inside the triangle, all the other three solutions had $P$ outside the triangle. The valid solution had $P = (9.5097811 , 5.657915)$

Answer 2

Observations and a conjecture:

In the figure below, $O$ is circumcenter of triangle $ABC$ and $K$ is centroid. $Q$ is the midpoint of $OK$, and hence lies on the Euler line.

Drawing $QF$, $QH$, $QJ$ perpendicular to $AC$, $BC$, $AB$, respectively, according to GeoGebra we have quadrilateral areas$$QJAF=\frac{QJ\cdot JA+AF\cdot FQ}{2}=170.94$$$$QFCH=\frac{QF\cdot FC+CH\cdot HQ}{2}=168.57$$$$QHBJ=\frac{QH\cdot HB+BJ\cdot JQ}{2}=166.84$$

These numbers are not far from their arithmetic mean of $168.78$. If the discrepancies are due only to rounding errors, then perpendiculars from point $Q$, midway between circumcenter and centroid, determine three equal quadrilaterals.

Moreover, for $\angle A >\angle B>\angle C$, if perpendiculars are instead dropped from $O$, then it can be seen that $OJAF$ is greatest in area and $OFCH$ is least, while if perpendiculars are dropped from $K$, then $KJAF$ is least in area and $KFCH$ is greatest. So again, equality requires an intermediate point like $Q$ or one very near it.