Geometry : what is the $\phi$ angle, if area of yellow rectangle is equal with area of red triangle?

Question

I have a right triangle and in it area of yellow rectangle is equal with area of red triangle. How could prove that $\phi=45^{\circ}$?

$$\text{Area of Yellow Rectangle}=\text{Area of Red Triangle}$$

This problem has no additional information. I'm sorry for the inconvenience.

Update:

If we add some constraint to this problem like bellow, is it have a unique solution? $$|FG|=x+y$$

Answer 1

Without additional information the angle $\varphi$ is not always $45^\circ$.

Let $x$ and $y$ be the sides of the rectangle and let $X$ and $Y$ be the legs of the right triangle. Then $xy=XY/2$ and $$\tan(\varphi)=\tan(\alpha-\beta)=\frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)} \tag{*}$$ where $\tan(\alpha)=(y+Y)/x$ and $\tan(\beta)=y/(x+X)$.

Now let $x=1$, $y=2$, $X=2$ and $Y=2$. Then $xy=XY/2$ but by (*) $$\tan(\varphi)=10/11\not=1=\tan(45^\circ).$$

On the other hand, as remarked by Quanto, if $x=y$ i.e. the yellow rectangle is a square then by (*) $$\tan(\varphi)=1=\tan(45^\circ).$$

Answer 2

In fact your issue has an infinity of solutions, making angle $\varphi$ dependent on position of point $M$ (see Fig. 1).

Indeed, one can build an infinity of such pairs (rectangle,triangle) with equal area (two examples are given on Fig. 1).

For every such pair, a specific angle (only exceptionally equal to 45°) is found.

Take a look at the following picture representing the locus of point $M(x,y)$ (common point to the rectangle and the triangle).

Fig. 1.

Remark : Let us find the equation of the locus of points $M(x,y)$.

As the area of rectangle with diagonal OM is $xy$ and the area of triangle $MNP$ is $\tfrac12(1-x-y)^2$, we get :

$$(1-x-y)^2=2xy \iff (x-1)^2+(y-1)^2=1,$$

i.e., is an arc of the circle centered in $(1,1)$ with radius $1$.

Fig. 2 : A funny Napoleon's hat : the way $\varphi$ varies (with a maximum of $\pi/4$) when $M$ varies on the arc of circle parametrized by $(x=1-\cos(\phi),y=1-\sin(\phi))$.

Answer 3

Thanks for all.

first equality: area of red and yellow are equal. $$ 2xy=vw \label{eq:1} \tag{1}$$

Now by using Thales' theorem in this triangle:

$$ \frac{v+y}{c}=\frac{b-x}{b} \Rightarrow v=c(\frac{b-x}{b})-y \label{eq:2} \tag{2}$$

$$ \frac{x+w}{b}=\frac{c-y}{c} \Rightarrow w=b(\frac{c-y}{c})-x \label{eq:3} \tag{3}$$

by importing $\ref{eq:2}$ and $\ref{eq:3}$ in $\ref{eq:1}$ and simplifying equations, I gained this equation:

$$ \frac{(x-b)^2}{b^2}+\frac{(y-c)^2}{c^2}=1 \label{eq:4} \tag{4}$$ It's means location of P, which area of red and yellow are equal, is an elliptic with canonical point $B$ and $C$. So we have $\infty$ points for $P$ in the $ABC$.

Now, How to find $\phi$?

By loking at this picture we have:

$$ AG=\sqrt{(x+w)^2+y^2} \label{eq:5} \tag{5}$$ $$ AF=\sqrt{(y+v)^2+x^2} \label{eq:6} \tag{6}$$ $$ FG=\sqrt{v^2+w^2} \label{eq:7} \tag{7}$$

Law of cosines

$$ FG^2=AG^2+AF^2-2AF.FG.\cos(\phi) \label{eq:8} \tag{8}$$

Now by using $\ref{eq:2}$ and $\ref{eq:3}$ in $\ref{eq:5}$, $\ref{eq:6}$ and $\ref{eq:7}$ and finally in $\ref{eq:8}$, and simplifying that, I reach to this equation:

$$ \sqrt{\left[\frac{b^2}{c^2}(c-y)^2+y^2 \right]\left[\frac{c^2}{b^2}(b-x)^2+x^2 \right]}\cos(\phi)= \left[bx+cy-xy(\frac{b^2+c^2}{bc})\right]\label{eq:9} \tag{9}$$

So

$$ \cos(\phi)=\frac{\left[bx+cy-xy(\frac{b^2+c^2}{bc})\right]}{\sqrt{\left[\frac{b^2}{c^2}(c-y)^2+y^2 \right]\left[\frac{c^2}{b^2}(b-x)^2+x^2 \right]}}\label{eq:10} \tag{10}$$

$$for \: b > x > 0 \: and \: c > y > 0 \: $$ For finding a $\phi$ as a function on $x$ and $y$ we need to considering $\ref{eq:4}$ and $\ref{eq:9}$. If I find exact solution I post it here.

I have writed a python code for determine all possible solution of this equation, but this not have a unique solution. Since all $x$ and $y$ that for them we have $\cos(\phi)=\frac{\sqrt{2}}{2}$ are solutions.