I have two circular arcs $ABCF$ and $ADEF$ which have the same endpoints $A, F$ and 'contain' the points $B, C$ and $D, E$ in the specified order.

How can I prove that there cannot be a circular arc $BCDE$, i.e. one that starts in $B$, intersects $C$ before $D$ and ends in $E$?

If both arcs are minor circular arcs, I can show that $B$ and $E$ must lie on different sides of the line through $C$ and $D$, hence either $BCD$ would be clockwise and $CDE$ counterclockwise or the other way around. Either way they could not form a circular arc $BCDE$ together.


However the case where one (or both) are major circular arcs cause me trouble.

Edit 1: Please note that the circular arcs may not overlap.

Edit 2: To clarify – the 'drawing' whose (non)existence I want to prove contains circular arcs $AB, BC, CF, AD, DE, EF, BC, CD, DE$. None of these may overlap and they shall form larger circular arcs together: $ABCF$ is made of $AB, BC, CF$; $ADEF$ is made of $AD, DE, EF$; and $BCDE$ is made of $BC, CD, DE$.


This answer calls figure 1 below "a quadrilateral $BCDE$" and figure 2 below "a quadrilateral $BCED$". The vertices of figure 1 are $B,C,D,E$ clockwise. On the other hand, the vertices of figure 2 are $B,C,E,D$ clockwise.

$\qquad\qquad\qquad$enter image description here

To prove by contradiction that there is no cirlce on which $B,C,D,E$ exist in this order, we suppose that there is such a circle.

Let us see, under the supposition, what conditions the four point $B,C,D,E$ in the figure given in the question has to satisfy.

Now, suppose that there is a circle on which $B,C,D,E$ exist in this order. (see the figure below. Note that the quadrilateral here is $BCDE$, not $BCED$.)

$\qquad\qquad\qquad$enter image description here

Then, from the inscribed angle theorem, we get $$\angle{BCE}=\angle{BDE}\tag1$$ $$\angle{CBD}=\angle{CED}\tag2$$ $$\angle{CDB}=\angle{CEB}\tag3$$ $$\angle{DCE}=\angle{DBE}\tag4$$

These are the necessary conditions which the four points $B,C,D,E$ in the figure given in the question has to satisfy to have a circle on which $B,C,D,E$ exist in this order.

Now let us come back to the figure given in the question.

We now know that the four points $B,C,E,D$ in the figure given in the question have to satisfy $(1)(2)(3)(4)$.

So, from $(1)(2)$, we have that the quadrilateral $BCED$ (not $BCDE$) has to be a parallelogram.

And from $(3)(4)$, the four inner angels have to be the same and so we have that the quadrilateral $BCED$ (again, not $BCDE$) has to be a rectangle. (see the figure below. The figure below is the same as the figure given in the question except that we have that the quadrilateral $BCED$ has to be a rectangle.)

$\qquad\qquad$enter image description here

Then, finally, we see that the order of the four points on the circumscribed circle of the rectangle $BCED$ has to be $B,C,E,D$.

This contradicts the supposition that there is a circle on which $B,C,D,E$ exist in this order.

It follows from this that there is no circle on which $B,C,D,E$ exist in this order. $\blacksquare$

  • $\begingroup$ How does it have to be a parallelogram? The thing you drew isn't even a parallelogram. $\endgroup$ – Christian Schnorr Feb 4 '17 at 14:15
  • $\begingroup$ @ChristianSchnorr: I wrote that the quadrilateral $BCED$ (note that it is not $BCDE$) is a parallelogram. I hope this helps. $\endgroup$ – mathlove Feb 4 '17 at 14:18
  • $\begingroup$ I don't see how that is a parallelogram either. Where are the parallel sides? $\endgroup$ – Christian Schnorr Feb 4 '17 at 14:19
  • $\begingroup$ @ChristianSchnorr: From $(1)$, we have $\angle{BCE}=\angle{BDE}$. Let $\alpha$ be this angle. From $(2)$, $\angle{CBD}=\angle{CED}$. Let $\beta$ be this angle. In the quadrilateral $BCED$, $2\alpha+2\beta=360^\circ$, so $\alpha+\beta=180^\circ$. From this, we see that $BC$ is parallel to $DE$ and that $CE$ is parallel to $BD$. $\endgroup$ – mathlove Feb 4 '17 at 14:24
  • $\begingroup$ I don't agree with (1)-(4) either. Just look at the drawing -- there are no parallel sides whatsoever. $\endgroup$ – Christian Schnorr Feb 4 '17 at 14:26

Suppose we have the unit circle centered at the origin. Then we can assign points $A$, $B$, $C$, $D$, $E$, and $F$ have coordinates $$A = \left( \cos \theta_1, \sin \theta_1 \right), $$ $$B = \left( \cos \theta_2, \sin \theta_2 \right), $$ $$C = \left( \cos \theta_3, \sin \theta_3 \right), $$ $$D = \left( \cos \theta_4, \sin \theta_4 \right), $$ $$E = \left( \cos \theta_5, \sin \theta_5 \right), $$ $$F = \left( \cos \theta_6, \sin \theta_6 \right), $$ where $$ \theta_1 < \theta_4 < \theta_5 < \theta_6 < \theta_3 < \theta_2.$$ Does this sort of reasoning take you any closer?

  • $\begingroup$ Not all 6 points need to lie on the same circle. If they did, then all three circular arcs $ABCF, ADEF, BCDE$ would be part of that very circle, and obviously the 'order' of points on the last circular arc would be violated. $\endgroup$ – Christian Schnorr Jan 31 '17 at 22:13

There is only one circle circumscribed circle which have for center the intercection of the bissection of the triangle.

The intersection of the bissections of a triangle is inside the Area of a triangle so it can not belong to both triangles !

  • $\begingroup$ I can't quite make the connection to bisections. Could you elaborate on that please? $\endgroup$ – Christian Schnorr Jan 31 '17 at 23:00
  • $\begingroup$ There is only one and only one circle passing through three non aligned points. We can know the center of this circle by finding the location of the circumcenter and If and only if a triangle is acute (all angles smaller than a right angle), the circumcenter lies inside the triangle. $\endgroup$ – user411953 Jan 31 '17 at 23:27
  • $\begingroup$ Therefore the cicumcenter of CEB and BED is not at the same location $\endgroup$ – user411953 Jan 31 '17 at 23:32
  • $\begingroup$ What guarantees that both CEB and BED are acute? $\endgroup$ – Christian Schnorr Jan 31 '17 at 23:34
  • $\begingroup$ If the two triangles are at the same time it seems that BCDE is a rectangle and they could form a circular arc, i guess i made a mistake... sorry $\endgroup$ – user411953 Jan 31 '17 at 23:44

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