# How to prove that no quadrilateral exist with three equal angles and increasing sidelengths

Working on an alternative solution to this recent question $\omega$ satisfies $a \omega^3 + b \omega^2 + c \omega + d = 0$, Prove that $|\omega| \leq \max( \frac{b}{a}, \frac{c}{b}, \frac{d}{c})$, I was brought to another question : how to prove that it is impossible for a quadrilateral $$PQRS$$ (see figure below) to have 3 equal angles $$a=\widehat{PQR}=\widehat{QRS}=\widehat{RSP}\tag{1}$$ and increasing sidelengths

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Fig. 1 : An obviously inexact figure (for the very fact that it is impossible to have such a quadrilateral !).

I have attempted to work in different directions, Bretschneider's formula, spirals, etc... without any result.

Remark : The connection with the cited problem or more exactly the formulation given in the fourth line of the accepted solution by @Jethro is as follows. Points $$P,Q,R,S$$ correspond resp. to complex numbers $$D,D+Cz,D+Cz+Bz^2,0$$ where $$z=re^{i\theta}$$ is a complex root of equation $$D+Cz+Bz^2+z^3=0$$ with common angle $$a=\pi-\alpha$$ and lengthes $$PQ=D,QR=Cr,RS=Br^2,SP=r^3$$. We have to prove that $$r \leq 1$$. Therefore, I have assumed that $$r>1$$, which, taking into account that $$D \leq C \leq B$$, yields relationship (2), aiming a contradiction.

• Omar Khayyam and Giovanni Saccheri proved many results similar to the one you want to prove, perhaps even including that one ... I will do some searching – Zubin Mukerjee Dec 7 '19 at 5:44
• @Zubin Mukerjee Thanks, I will be waiting... – Jean Marie Dec 7 '19 at 5:47
• Not a complete answer yet, just my thoughts so far, so I will comment here:$$\,$$Suppose for contradiction that there is such a quadrilateral.$$\,$$ Case 1: $$0 < a < \pi/2$$ Case 2: $$a = \pi/2$$ Case 3: $$\pi/2 < a < 2\pi/3$$ Case 4: $$2\pi/3 < a$$ Clearly Case 2 is just a square, so this doesn't work. Case 4 violates angle sum of a quadrilateral. For Case 1: Let $A$ be the intersection of $\overleftrightarrow{QP}$ and $\overleftrightarrow{RS}$. Let $B$ be the intersection of $\overleftrightarrow{SP}$ and $\overleftrightarrow{RQ}$. Then $$\triangle ARQ \sim \triangle BSR$$ – Zubin Mukerjee Dec 7 '19 at 6:17

As shown in the diagram above, join $$Q$$ to $$S$$. Also, let $$\angle SQP = d$$, so $$\angle SQR = a - d$$, plus let $$\angle QSR = f$$, so $$\angle QSP = a - f$$.

As shown & explained in Relationship of side lengths and angles of a triangle, plus as mentioned in the comment about Euclid's I.18, the angle opposite the greater side length is greater. Thus, with $$\triangle QSR$$, since $$SR \gt QR$$, you have

$$\angle SQR \gt \angle QSR \implies a - d \gt f \tag{1}\label{eq1A}$$

Next, with $$\triangle QSP$$, since $$SP \gt PQ$$, you have

$$\angle SQP \gt \angle QSP \implies d \gt a - f \implies - a + d \gt -f \implies a - d \lt f \tag{2}\label{eq2A}$$

However, this contradicts \eqref{eq1A} (also, adding $$a - d \gt f$$ to $$d \gt a - f$$ gives $$a \gt a$$ for another way to see this doesn't work). This means the stated conditions of the relationships of the side lengths, i.e., $$PQ \lt QR \lt RS \lt SP$$, cannot be correct. Note, also, this proof doesn't even need, or use, that $$\angle QRS = a$$ or $$RS \gt SP$$.

• Here you are using Euclid's I.18? "In any triangle the angle opposite the greater side is greater." – Zubin Mukerjee Dec 7 '19 at 6:11
• @ZubinMukerjee You are correct. I just provided a link to a Web page which also explains this, and will add your link as well. – John Omielan Dec 7 '19 at 6:13
• Is there a typo on $(1)$? Both read $\angle QSR$ .. maybe should be this? $$\angle SQR > \angle QSR$$ – Zubin Mukerjee Dec 7 '19 at 6:21
• OK, looks good now. Because this only uses Euclid's I.18, and you never used the parallel postulate, I think you have proved that such a quadrilateral cannot exist in Euclidean geometry, nor in hyperbolic geometry ... very nice, +1 :) – Zubin Mukerjee Dec 7 '19 at 6:28
• @ZubinMukerjee Thanks for the compliment. To help make it easier to visualize, I've just added a rough diagram. Note I was only interested in proving this for Euclidean geometry, so if it also works in hyperbolic geometry, that is a definite bonus. – John Omielan Dec 7 '19 at 6:37

Using an aspect of the Inscribed Angle Theorem that tells us $$Q$$ and $$S$$ must lie on congruent circular arcs with common chord $$\overline{PR}$$, we have this diagrammatic proof:

Like @JohnOmielan's proof, this one doesn't use any information about $$\angle R$$. Unlike @John's proof, this one is only valid in Euclidean geometry (where IAT holds). It's worth noting that both proofs are valid even for self-intersecting $$\square PQRS$$ with $$Q$$ and $$S$$ on the same side of $$\overline{PR}$$.

Regarding a hyperbolic counterpart: The locus of a point that makes a constant angle with two other points is a simple curve. According to this answer on MathOverflow, the locus in the Klein model is actually a circle. So, embedding my diagram in the Klein disk with $$M$$ at the origin almost works. One would still need to show that smaller arcs (in the model) correspond to smaller lengths (in the geometry); presumably, the argument would effectively duplicate @John's logic, so this diagrammatic approach merely adds unnecessary complication.

• A simple proof indeed ! – Jean Marie Dec 7 '19 at 10:54