Divide a right triangle into three quadrilaterals of equal area

Question

The goal is to find the point $M$ inside a given right triangle $ABC$ such that $\operatorname{Area}(APMN)=\operatorname{Area}(CQMP)=\operatorname{Area}(BQMN)$, where $N$, $P$, and $Q$ are the orthogonal projections of $M$ onto $\overline{AB}$, $\overline{AC}$, and $\overline{BC}$, respectively, given that $\lvert AB\rvert=5$ and $\lvert BC\rvert=11$.

To do this, I worked in the "natural" coordinate system with origin $A$ and considering $M(x;y)$. So: $3xy = \operatorname{Area}(ABC)$, and I thus found the coordinates of $Q$ and then $\operatorname{Area}(CQMP)$, for example, as a function of $x$ and $y$.

This led to the equation $(x-2)\left(3x^3-24x^2-190x+100\right)=0$.

But I wonder if another method is possible, whether synthetic or by using the Pythagorean theorem, because I have the feeling that this problem that I imagined must have been known even before Descartes!?

Answer 1

Here is a generic quartic equation using a different approach.

Consider a right $\triangle ABC$ with legs $BC=a$ and $AB=b$ (as shown in the diagram below).

Let $(x,y)$ be the coordinates of point $M$ lying inside $\triangle ABC$ such that it divides $\triangle ABC$ into three equal areas $PMQB$ (labelled 'grey'), $[MRCQ]$ (labelled 'blue') and $[ARMP]$ (labelled 'yellow'). $$\therefore [PMQB]=[MRCQ]=[ARMP]=\frac{1}{3}[\triangle ABC]\tag 1$$ From $[PMQB]=\frac{1}{3}[\triangle ABC]$, $$xy=\dfrac13\left(\dfrac{1}{2}ab\right)\implies y=\dfrac{ab}{6x}\tag 2$$ From Eq.(1), total area of $\triangle ABC$, $$[PMQB]+[MRCQ]+[ARMP]=[\triangle ABC]$$ $$xy+([\triangle APM]+[\triangle ARM])+([\triangle MQC]+[\triangle MRC])=\dfrac 12ab$$ $$xy+[\triangle APM]+[\triangle MQC]+[\triangle AMC]=\dfrac 12ab\ \ \ \quad \left(\because [\triangle ARM]+[\triangle MRC]=[\triangle AMC]\right)$$

$$xy+\dfrac12(b-y)x+\dfrac 12(a-x)y+\dfrac12(MR)\sqrt{a^2+b^2}=\dfrac 12ab$$ $$\implies MR=\dfrac{ab-bx-ay}{\sqrt{a^2+b^2}}\tag 3$$ Now from $\triangle APS\sim \triangle ABC$, $$\dfrac{PS}{BC}=\dfrac{AP}{AB}\implies PS=\dfrac{a}{b}(b-y)$$ From Eq.(1), $$[PMQB]+[MRCQ]=\frac{2}{3}[\triangle ABC]$$ $$[PSCB]+[\triangle MRS]=\frac{2}{3}[\triangle ABC]$$ $$\dfrac{1}{2}(PS+BC)(MQ)+\dfrac{1}{2}(MR)(RS)=\frac{2}{3}\left(\dfrac 12ab\right)$$ $$\dfrac{1}{2}\left(\dfrac{a}{b}(b-y)+a\right)y+\dfrac{1}{2}(MR)(MR\tan\alpha)=\dfrac 13ab$$ $$\dfrac{1}{2}\left(\dfrac{2ab-ay}{b}\right)y+\dfrac{1}{2}(MR)^2\dfrac{a}{b}=\dfrac 13ab$$ $$\dfrac{2aby-ay^2}{2b}+\dfrac{a}{2b}\left(\dfrac{ab-bx-ay}{\sqrt{a^2+b^2}}\right)^2=\dfrac 13ab\quad \quad \left(\text{substituting MR from}\ Eq.(3)\right)$$ $$\dfrac{2ab\cdot \dfrac{ab}{6x}-a\left(\dfrac{ab}{6x}\right)^2}{2b}+\dfrac{a}{2b}\left(\dfrac{ab-bx-a\cdot \dfrac{ab}{6x}}{\sqrt{a^2+b^2}}\right)^2=\dfrac 13ab\quad \quad \left(\text{substituting y from}\ Eq.(1)\right)$$ $$36x^4-72ax^3+24(a^2-b^2)x^2+12ab^2x-a^2b^2=0\tag 4$$ The above quartic equation (4) always gives a unique positive real value of $x$, and the corresponding value of $y$ from Eq.(2), such that the point $M(x,y)$ lies inside the right $\triangle ABC$ and hence it always satisfies the following condition $$a, b, x, y, ab-bx-ay>0\iff x\in \left(\dfrac{(3-\sqrt{3})a}{6}, \dfrac{(3+\sqrt{3})a}{6}\right) $$ In general, for a right $\triangle ABC $ with its legs $BC=a$, and $AB=b$ coinciding with the +ve x and y-axes respectively, the coordinates of point $M$ inside triangle are given as $$\boxed{\color{blue}{M(x,y)\equiv\left(x,\dfrac{ab}{6x}\right)}}$$ where $x$ is a root of the following quartic equation: $$\boxed{\color{blue}{36x^4-72ax^3+24(a^2-b^2)x^2+12ab^2x-a^2b^2=0\quad \forall \ x\in \left(\dfrac{(3-\sqrt{3})a}{6}, \dfrac{(3+\sqrt{3})a}{6}\right) }}$$

Now, a per given values in the question, $a=BC=5$ and $b=AB=\sqrt{11^2-5^2}=4\sqrt{6}$, the quartic equation is as follows $$36x^4-72(5)x^3+24(5^2-(4\sqrt6)^2)x^2+12(5)(4\sqrt6)^2x-(5)^2(4\sqrt6)^2=0$$ $$3x^4-30x^3-142x^2+480x-200=0\tag 5$$ $$(x-2)(3x^3-24x^2-190x+100)=0$$ Eq(5) is the same quartic equation as obtained by OP that leads to the same solution as given by @Intelligenti pauca.

Answer 2

Setting: $$ a=BC=11,\quad c=AB=5,\quad b=AC=\sqrt{a^2-c^2},\\ x=MP,\quad y=MN,\quad z=MQ, $$ we have the equations: $$ \cases{ \begin{align} &xy={1\over6}bc\\ &bx+cy+az=bc\\ &(b-y)x+z\sqrt{(b-y)^2+x^2-z^2}={1\over3}bc \end{align}} $$ This is an 8th-degree system. Its only positive solution is: $$x=2,\quad y=5\sqrt{2\over3},\quad z=\sqrt{2\over3}$$