Construct quadrangle with given angles and perpendicular diagonals

Solution 1:

Consider a hypothetical solution $\square ABCD$ with diagonals meeting at $X$, and with angle measures and segment lengths as shown:

Then

$$\tan \alpha = \tan A = \tan(\alpha_1 + \alpha_2) = \frac{\tan\alpha_1+\tan\alpha_2}{1-\tan\alpha_1\tan\alpha_2} = \frac{\frac{d}{a}+\frac{b}{a}}{1-\frac{d}{a}\frac{b}{a}} = \frac{a(b+d)}{a^2-bd} \tag{1}$$ $$\tan\beta = \frac{b(a+c)}{b^2-ac} \tag{2}$$ $$\tan\gamma = \frac{c(b+d)}{c^2-bd} \tag{3}$$ $$\tan\delta = \frac{d(a+c)}{d^2-ac} \tag{4}$$

To show that this hypothetical solution is valid, we need only solve equations (1) through (4) for $b$, $c$, $d$ in terms of $\alpha$, $\beta$, $\gamma$, $\delta$, and $a$ (which we can take to be $1$). This is do-able, and the algebra gets no more complicated than quadratics (so that the solution is constructible), but the expressions are a bit messy. I'll post more after I do some clean-up.

Edit. After considerable manipulation, (I think) the above equations reduce to these: $$\begin{align} (a^2-b^2)\sin\alpha\sin\beta \cos(\gamma+\beta) + a b \left( \sin\alpha \cos(\gamma+2\beta) + \cos\alpha\sin\gamma \right) &= 0 \\[4pt] (a^2-d^2)\sin\alpha\sin\delta \cos(\gamma+\delta)\;+ a d \left( \sin\alpha \cos(\gamma+2\delta) + \cos\alpha\sin\gamma \right) &= 0 \\[4pt] 2 (a^2+c^2) \sin\alpha\sin\gamma \cos(\gamma+\beta) \cos(\gamma+\delta) \qquad- a c ( k + \sin^2(\alpha-\gamma) ) &= 0 \end{align}$$ where $$k := -1 + \sin^2\alpha + \sin^2\beta + \sin^2\gamma + \sin^2\delta + \cos(\alpha-\gamma) \cos(\beta-\delta)$$

Consequently, we have $$\begin{align} \frac{b}{a} &= -\frac{ \sin\alpha \cos(\gamma+2\beta) + \cos\alpha \sin\gamma \pm \sqrt{k}}{2\sin\alpha \sin\beta \cos(\gamma+\beta)} \\[6pt] \frac{d}{a} &= -\frac{ \sin\alpha \cos(\gamma+2\delta) + \cos\alpha \sin\gamma \pm \sqrt{k}}{2\sin\alpha \sin\delta \cos(\gamma+\delta)} \\[6pt] \frac{c}{a} &= \frac{ k + \sin^2(\alpha-\gamma) \pm 2 \sin(\alpha-\gamma) \sqrt{k}}{4 \sin\alpha \sin\gamma \cos(\gamma+\beta)\cos(\gamma+\delta)}= \frac{\left(\;\sin(\alpha-\gamma) \pm \sqrt{k} \;\right)^2 }{4 \sin\alpha \sin\gamma \cos(\gamma+\beta)\cos(\gamma+\delta)} \end{align}$$ where, for now at least, resolution of the "$\pm$"s is left as an exercise to the reader.

Note that the relation $\alpha+ \beta+\gamma+\delta = 360^\circ$ causes each of these to have myriad representations. I don't claim to have given the best possible ones; in fact, I suspect there are representations that make the relations far more clear.

As mentioned, the various quantities are constructible, since the lengths are at most as complicated as a square root. Formulation of a construction strategy will have to wait.

Edit 2. If we normalize our lengths, say, with a constant sum, $$a + b + c + d = s$$ then we can express each length independently. With $m := \pm\sqrt{k}$, we have $$\frac{a}{s} = \frac{\left(\; m + \sin(\alpha-\gamma) \;\right)\left(\;m + \sin(\beta+\delta) + 2 \sin\beta\sin\delta\;\right)}{2m\left(\;2\sin(\beta+\delta)+\cos(\beta-\delta)-\cos(\alpha-\gamma) \;\right)}$$ while expressions for $b$, $c$, $d$ arise by cyclically permuting the angles, $\alpha\to\beta\to\gamma\to\delta\to\alpha$. A different normalization (for instance, $a^2+b^2+c^2+d^2=s^2$ seems a classic choice) would lead to different —potentially better— representations, but my attempts at symbol-wrangling haven't resulted in anything particularly nice.

By the way, to verify that the earlier ratios hold, it helps to know that $$m^2 - \sin^2(\alpha-\gamma) = 4\sin\alpha\sin\gamma\cos(\gamma+\beta)\cos(\gamma+\delta)$$ Therefore multiplying $a$ by the $c/a$ ratio above turns out to be an overly-complicated way to flip a single sign: $m + \sin(\alpha-\gamma) \;\to\; m - \sin(\alpha-\gamma)$, which matches the considerably-easier process of exchanging $\alpha$ and $\gamma$ (and exchanging $\beta$ and $\delta$, which actually does nothing) in the formula for $a$.

Solution 2:

The problem posted is equivalent to this one:

Let $a'$, $b'$, $c'$ and $d'$ rays starting from point $M$, such that $m(\angle a'Mb')=\alpha$, $m(\angle b'Mc')=\beta$, $m(\angle c'Md')=\gamma$ and $m(\angle d'Ma')=\delta$. Construct a rectangle $FGHI$ such that $F \in b'$, $G \in c'$, $H\in d'$ and $I \in a'$. See fig. 1.

Fig.1 - Rectangle that develops to the required quadrangle.

The reason: If we reflect point M across the lines $f$, $g$, $h$ and $i$ (the sides of the rectangle), we will get the quandrangle required by OP.

To solve this new problem, let's derive some equations. See fig. 2.

Fig.2 - Rectangle in a coordinate system.

Let $$I=(-1, \tan \alpha),$$ $$F=(-b, 0),$$ $$G=(a, a \tan \beta).$$ As line $i$ is perpendicular to line $f$, we get: $$a+b=\frac{a \tan \alpha \tan \beta}{1-b} \quad (1),$$ or $$a=\frac{b(1-b)}{ \tan \alpha \tan \beta -1+b} \quad (2).$$ As $x_H-x_I=x_G-x_F$, we get: $$\tan(\alpha+\delta)- \tan \alpha=(a+b) \tan(\alpha+\delta) +a \tan \beta. \quad(3)$$ Substituting $(1)$ and $(2)$ in $(3)$, we get: $$\tan \beta b^2 + [\tan(\alpha+\delta)- \tan \alpha-\tan\alpha \tan \beta \tan(\alpha+\delta)- \tan \beta]b+ [\tan(\alpha+\delta)- \tan \alpha](\tan\alpha \tan \beta-1)=0. \quad (4)$$ Lill's method is apropriated to solve eqution $(4)$ for $b$. The coefficients of equation $(4)$ can be easily constructed, since they are products or the results of additions/subtractions of constructible quantities.

Knowing the value of $b$ the construction of the rectangle and of the needed reflections is not difficult.