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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

$$0<PQ<QR<RS<SP\tag{2}$$

enter image description here

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.

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    $\begingroup$ 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 $\endgroup$ – Zubin Mukerjee Dec 7 '19 at 5:44
  • $\begingroup$ @Zubin Mukerjee Thanks, I will be waiting... $\endgroup$ – Jean Marie Dec 7 '19 at 5:47
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    $\begingroup$ 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 $$\triangle ARQ \sim \triangle BSR$$ $\endgroup$ – Zubin Mukerjee Dec 7 '19 at 6:17
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Quadrilateral with 3 equal angles of a

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 $RS \gt SP$.

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    $\begingroup$ Here you are using Euclid's I.18? "In any triangle the angle opposite the greater side is greater." $\endgroup$ – Zubin Mukerjee Dec 7 '19 at 6:11
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    $\begingroup$ @ZubinMukerjee You are correct. I just provided a link to a Web page which also explains this, and will add your link as well. $\endgroup$ – John Omielan Dec 7 '19 at 6:13
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    $\begingroup$ Is there a typo on $(1)$? Both read $\angle QSR$ .. maybe should be this? $$\angle SQR > \angle QSR$$ $\endgroup$ – Zubin Mukerjee Dec 7 '19 at 6:21
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    $\begingroup$ 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 :) $\endgroup$ – Zubin Mukerjee Dec 7 '19 at 6:28
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    $\begingroup$ @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. $\endgroup$ – John Omielan Dec 7 '19 at 6:37
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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:

enter image description here

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.

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    $\begingroup$ A simple proof indeed ! $\endgroup$ – Jean Marie Dec 7 '19 at 10:54

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