Splitting rectangle of area 1 into three equal triangles

Question

As shown in the figure below, the area of rectangle $ABCD = 1$, $Q$ and $P$ are points on $CD$ and $BC$.
Such that $\triangle ADQ = \triangle QCP = \triangle ABP$
Find the area of $\triangle ADQ$.

I've tried assuming the three triangles' area is $a$ each, $AD=x, AB=y$, then figuring out $DQ, BP$ by areas of $\triangle ADQ, \triangle ABP$, then making an equation for the area of $\triangle QPC$.

Only knowing $xy=1$, my method doesn't seem to work very well. I get $x^2+y^2$ terms at the end. Is there a way to find $x^2+y^2$?

Answer 1

Following the OP's notation for areas and lengths, we find $DQ=\dfrac{2a}x$ and $BP=\dfrac{2a}y$.

Therefore $PC=x-\dfrac{2a}y$ and $QC=y-\dfrac{2a}x.$

Now considering the area of $\triangle PCQ$,

$$a=\frac12\left(x-\frac{2a}y\right)\left(y-\frac{2a}x\right)$$ $$2a=xy-4a+\frac{4a^2}{xy}$$ $$4a^2-6a+1=0$$

(because $xy=1$)

And $a$ is the area required. Can you take it?

Answer 2

HINT.

Set $AD=a$, $CD=1/a$, $DQ=x$, $BP=y$. Equating the areas you'll end up with a single quadratic equation for $(ax)$.

Answer 3

It is given that $AD=x$, $AB=y$, and $xy=1$. Let's denote $CQ=a$ and $BP=b$. Now, we can express the areas of the three right-angled triangles $APB$, $QCP$, and $ADQ$ in terms of $x$, $y$, $a$, and $b$ as shown below. $$\triangle APB =\dfrac{yb}{2} \tag{1}$$ $$\qquad\space\triangle QCP=\dfrac{ax-ab}{2} \tag{2}$$ $$\qquad\qquad\qquad\quad\triangle ADQ=\dfrac{xy-xa}{2}=\dfrac{1-xa}{2}\tag{3} $$

Since the three mentioned triangles have the same area, we shall write, first using (1) and (2), $$a=\dfrac{yb}{x-b}, \tag{4}$$ and then using (2) and (3), $$a\left(2x-b\right)=1. \tag{5}$$

when we substitute the value of $a$ from (4) in (5), we get, $$\dfrac{yb}{x-b}\left(2x-b\right)=1 \quad\longrightarrow\quad 2xyb-yb^2=x-b \quad\longrightarrow\quad yb^2-3b-x=0. $$

The roots of the quadratic equation in $b$ are, $$b=\dfrac{3\pm\sqrt{9-4xy}}{2y}=\dfrac{3\pm\sqrt{5}}{2y}=\left(\dfrac{3\pm\sqrt{5}}{2}\right)x.$$

Since $\dfrac{3+\sqrt{5}}{2}\gt 1$, $b$ has only one useful root, i.e., $$b=\left(\dfrac{3-\sqrt{5}}{2}\right)x. \tag{6}$$

Therefore, the area of any of the three triangles can be expressed using (1) as, $$\triangle = \dfrac{yb}{2}=\left(\dfrac{3-\sqrt{5}}{4}\right)xy=\left(\dfrac{3-\sqrt{5}}{4}\right)\approx 0.190983.$$

Answer 4

Let $|AB|=x, |AD|=y, |BP|=a, |DQ|=b$.

As the areas of $\triangle ABP$ and $\triangle ADQ$ are equal, we have $xa=yb$, or

$$\frac{x}{y}=\frac{b}{a}$$.

This means the point $(Q_x,P_y)$ lies on $AC$, and that $a=ky, b=kx$.

Twice the area of $\triangle QDB$ is therefore $(y-kx)(x-ky)=xy(1+k^2)-k(x^2+y^2)$, and as it is equal to the other two, equals $k$.

As $xy=1$, solve

$$1-k(x^2+y^2+1)+k^2=0$$

where $x^2+y^2$ is particular to the given rectangle $ABCD$, and so

$$k=\frac{-1\pm\sqrt{(x^2+y^2+1)^2-4}}{2}$$

If $x=y=1$, this gives the golden ratio.

The final area of $\triangle APQ$ is then $1-\frac{3k}{2}$.