The following came up when I worked on the answer for a different question (though it was ultimately not used in this form):

Proposition. Given positive angles $\alpha,\beta,\gamma,\delta$ with $\alpha+\beta+\gamma+\delta=360^\circ$, $\beta<180^\circ$, $180^\circ< \alpha+\beta<270^\circ$, $180^\circ< \beta+\gamma<270^\circ$, there exists a convex quadrangle $ABCD$ with $\angle A=\alpha$, $\angle B=\beta$, $\angle C=\gamma$, $\angle D=\delta$, having $AC\perp BD$.

Proof: Ignoring the orthognality condition, there are many possible quadrangles with the given angles that can be continuously transformed into each other. For such let $P$ denote the intersection of $AC$ and $BD$. In the degenerate case $A=B$, we get $\angle CPD=\alpha+\beta-180^\circ<90^\circ$, in the degenerate case $B=C$, we get $\angle DPA=\beta+\gamma-180^\circ<90^\circ$. enter image description here Then the Intermediate Value Theorem guarantees the existence of a case where $\angle CPD=90^\circ$. $_\square$

My question is: Can someone provide a proof not relying on continuity arguments? That is, something more classic Greek compass-and-straightedge-y constructive?

Edit: I had to update and add $\beta<180^\circ$ to the condition in the proposition - the old version would have allowed $\beta\ge180^\circ$ and so no convex quadrilateral at all. If we allow nonconvex quadrangles and diagonals to interesect in the exterior, this additional condition should be unnecessary.

  • $\begingroup$ nice problem. this is just an idea i have been thinking about. have you tried inversion on a circle with one side as the diameter? $\endgroup$ – abel Dec 10 '14 at 13:24
  • $\begingroup$ @abel not with success $\endgroup$ – Hagen von Eitzen Dec 12 '14 at 9:34
  • $\begingroup$ I have a question. Do you want a descriptive method to draw a quadrilateral given the angles and perpendicular diagonals using basic instruments? $\endgroup$ – Alex Silva Dec 12 '14 at 15:59
  • $\begingroup$ For a moment, I hoped that given two rays at specified angles from a given segment (the rays $\vec{CB}$ and $\vec{DA}$ from $CD$ in your figure), the possible segments $AB$ that result in perpendicular diagonals would all intersect at one constructable point. Had they, you'd be able to find the one that makes the correct angle for your input. Unfortunately, they don't, but the locus of such sides $AB$ makes a nice picture. I don't know if it's any help to think of the problem this way, but here's the picture: i.imgur.com/XeZG0T1.gif $\endgroup$ – Steve Kass Dec 13 '14 at 17:55
  • $\begingroup$ Deleted my answer because it seems I've completely misunderstood the question. Sorry. $\endgroup$ – Han de Bruijn Dec 13 '14 at 18:12

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

enter image description here


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

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 $a$.

  • $\begingroup$ Isn't the condition $\alpha+ \beta+\gamma+\delta = 360^\circ$ just redundant? Given that we are searching for a quadrangle with those angles. $\endgroup$ – Han de Bruijn Dec 13 '14 at 10:14
  • $\begingroup$ @HandeBruijn: Redundant, perhaps, but not irrelevant. When verifying my work, or seeking better representations for the values, or devising a construction, it's worth remembering that I could've written, say, $\cos(\alpha+\delta)$ for $\cos(\beta+\gamma)$, and so forth. $\endgroup$ – Blue Dec 13 '14 at 11:25
  • $\begingroup$ If we decompse $ \pi/2 = \alpha_2 + \beta_1 , \pi/2= \beta_2 + \gamma_1 ,..,...$ with the given equations do we not get bounds of each plausible component? $\endgroup$ – Narasimham Dec 13 '14 at 20:30
  • $\begingroup$ @Narasimham: Have been trying just that, without getting any further. But anyway the condition $\alpha+\beta+\gamma+\delta=2\pi$ is automatically fulfilled then. $\endgroup$ – Han de Bruijn Dec 14 '14 at 16:14
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    $\begingroup$ I'll accept this as it clearly demonstrates constructability and it seems that an explicit construction (at least one straightforwardly derived from these expressions) is a task for long winter nights ;) $\endgroup$ – Hagen von Eitzen Dec 15 '14 at 14:13

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. RectangleToQuadriangle

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.

enter image description here

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.

  • $\begingroup$ A very interesting twist on the problem statement! $\endgroup$ – Hagen von Eitzen Dec 18 '14 at 11:36
  • $\begingroup$ Lill's method? Why so difficult? Isn't that equation (4) just a quadratic one? $\endgroup$ – Han de Bruijn Dec 18 '14 at 12:08
  • $\begingroup$ @HandeBruijn Since the OP wanted a compass-and-straightedge construction I thought it would be easier use equation(4) as it is. But it is possible to use another method (a graphical or algebraic one) to solve equation (4). $\endgroup$ – RicardoCruz Dec 18 '14 at 12:21

I think ruler compass construction should be completed at first;

[Earlier, redone.The construction can be completed only upto two parameters of arbitrariness , one for scale ( naturally as only angles $ \alpha,\beta, \gamma$ and $\delta $ are given) and another for adjoining sides ratio AB/BD at B.Draw AB and AD so that angle DAB = $ \alpha$. Construct semi circle on AB and AD as diameters intersecting at point O, where yet un-constructed diagonals would intersect at right angles.Join OB,OD, OA and extend AO on the other side towards an yet unknown point/direction C.Mark off $ \beta$ and $ \delta $ at at B and D respectively. The three lines are concurrent at C making angle $ 2\pi -( \alpha+\beta+\delta) $ A Java or Geometric SketchPad can track all solutions by changing AB/BD ratio or parameter with mouse. A static sketch would be soon uploaded. It is not necessary as an answer but for labeling. EDIT1: Regarding parametric variation, shall add a detail missing here].


My approach is partially changed as under, only one circle is needed.

Rotating CD Line

Angles at 1 and 2 are right angles in a semi-circle. When AB is fixed and lines BC,BD are drawn, the angle sum $ \gamma + \delta $ is a given constant. (In either position 1 or position 2 angle sum of Line L1 or L2 is the same). The inclination of variable Line L is a trig function of $ \theta $, $ S(\theta) $, which is to be worked out. For correct $\theta$ position point on semi-circle solved for between 1 and 2, Line L becomes parallel to CD.Slope of L should match that of CD.

Hope someone takes it from there, I have to go.


Any solution can be stretched/shrunken and still fit. So you can choose one length, e.g. the diagonal AC.

The following is much easier to understand if you draw as you read (I don't know how to easily provide drawings myself).

My idea is to draw AC (I would do it horizontally, C being right from A). Then angles $\beta$ and $\delta$ will give you arches that contain all possible positions of points B (above) and D (below); both arches start in A and end in C. (If one of the two angles is greater than $180^∘$ then that arch will be on the other side, same as the other arch.)

Now pick any line perpendicular to AC that intersects both arches. Wherever it intersects them, they are B' and D'. (It may intersect in 2 places, in which case you have 2 possibilities for B' and D'.) If the angles $\alpha$ and $\gamma$ are right (they are either both right or both wrong), then, of course, B = B' and D = D'.

Here's where I'm a bit stuck with the construction: how to pick the right perpendicular line for the angle $\alpha$? Too far left and the angle will be to big, too far left and the angle will be too small. Obviously for $\alpha=180^∘$ the intersecting line should go through A, for bigger somewhere left from A, for smaller somewhere right from A.

  • $\begingroup$ Nice idea, but this looks like just a different way of making use of the Intermediate Value Theorem. At least you made me aware that I should add a condition to the proposition statement if I really want to ensure a convex quadrilateral. $\endgroup$ – Hagen von Eitzen Dec 8 '14 at 17:22
  • $\begingroup$ @HagenvonEitzen I don't think that makes much difference to the difficulty of construction though. $\endgroup$ – Heimdall Dec 8 '14 at 18:49

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