Talk:Bicentric quadrilateral

Fuss' Theorem

Do all positive triplets (x, r, R) satisfying Fuss' equation have an associated bicentric quadrilateral? It seems to me that they don't necessarily have one. For example, let r be very close to zero, so by Fuss' formula x is very close to R. Then we have a tiny inner circle very near to the circumference of the outer circle, and it would appear to be impossible to draw a bicentric quadrilateral in this case. So are there some inequality restrictions that have to be imposed on r, x, and R along with Fuss' equation in order to guarantee existence? Duoduoduo (talk) 21:00, 17 August 2011 (UTC)

I am almost sure that I have read somewhere that Fuss' theorem is also a sufficient condition for the existence of a bicentric quadrilateral given circles of radii r, R and center distance x, but I cannot recall a reference. The situation you describe is possible. You can experiment on it using a dynamic geometry computer program like GeoGebra (free). Then you find that in your situation, the angle between two of the sides in the bicentric quadrilateral (those between the two close circles of very different sizes) is very close to $\pi$, so the bicentric quadrilateral is almost a triangle.

There is also another possibility. Solving Fuss' equation for x gives two solutions since it is a quadratic equation. The solution I wrote (with minus in front of the inner radical) is for bicentric quadrilaterals. The other solution is for cyclic quadrilaterals with an excircle (sometimes denoted escribed circle). In fact, I have been thinking about starting a new article on quadrilaterals with an excircle. These are harder to find info about, but they have a few very closely related properties to tangential ones. Circlesareround (talk) 17:29, 18 August 2011 (UTC)

See Ex-tangential quadrilateral for the new article on quadrilaterals with an excircle. Circlesareround (talk) 20:34, 18 August 2011 (UTC)

While I haven't tried a dynamic geometry program, I'm still skeptical that it can be done in the case I described. I recently added the passage "If two circles, one within the other, are the incircle and the circumcircle of a bicentric quadrilateral, then every point on the circumcircle is the vertex of a bicentric quadrilateral having the same incircle and circumcircle. [10]", which assertion I got from Mathworld. So with the inner circle very near the rightmost part of the outer circle, pick a vertex on the upper left portion of the outer circle. Draw a tangent to one side of the small circle to a second vertex, and draw from the original vertex another tangent to the other side of the small circle to get a third vertex. Then in the small space between the circles, it seems impossible to draw two more tangents to complete the quadrilateral. Duoduoduo (talk) 23:31, 18 August 2011 (UTC)

It works!

This is a very interesting problem. I have made a figure in GeoGebra to illustrate and hopfully convince you that it is possible. I drew a bicentric quadrilateral ABCD with contact quadrilateral WXYZ (you start the construction here). Then as you suggested, I took a random point to the upper left. It is E. From this point tangents to the incircle is constructed to get F and G. From G the tangent to the incircle meets the circumcircle exactly at E, closing the quadrilateral as Poncelet's closure theorem describes. Now, my incircle is not very tiny, since then you would not see anything important at the screen if you cannot move around and magnify. This can be done in GeoGebra and it works fine also. Circlesareround (talk) 16:03, 19 August 2011 (UTC)

Okay, I'm pretty much convinced. Still, it would be good if you could find a source for an "if and only if" statement to go with Fuss' equation, so that could be put into the article.

Also, you are obviously good with imaging -- could you redo the image you put on this discussion page, moving the inner circle farther away from the outer circle for visual clarity, and put the revised image at the end of the "Other properties" section where it says that any point on the circumcircle could be a vertex? Duoduoduo (talk) 16:45, 19 August 2011 (UTC)

The reason I have reworded the statement of the converse of Fuss's Theorem is that I had a hard time with the passage

Fuss' theorem is ... a necessary condition for a quadrilateral to be bicentric....

The parameters in Fuss's condition are not defined unless the quadrilateral is bicentric, so it seems to me that there's something circular in that statement.

Nice find on the 1909 source, Circlesareround ! Duoduoduo (talk) 16:54, 30 September 2011 (UTC)

construction - recent addition

I removed the following addition as to very least it needs some work before it maybe included

removed

A bicentric quadrilateral $A_{1}A_{2}A_{3}A_{4}$ according to J. L. Coolidge
when $r\;{\widehat {=}}\;R$ and $\rho \;{\widehat {=}}\;r$ then is
${\frac {1}{(R+d)^{2}}}+{\frac {1}{(R-d)^{2}}}={\frac {1}{r^{2}}}$

Another possibility is shown by Julian Lowell Coolidge, he published it in 1916 in his book A Treatise on the Circle and the Sphere a construction of the convex quadrilateral.

"If two circles be so related that a triangle or quadrilateral may be inscribel in the one and circumscribed to the other, then an infinite number of such triangles or quadrilaterals may be found, one vertex being taken at random on the other circle."

Since the diagonal A₁ A₃ passes through the two centers O' and O of the circles K_r and K_R, it is also a right kite. The following construction uses the basic description contained in his book as well as the relevant figure as a basis.

It begins with the drawing of the circle K_R around the central point O with the freely selected radius R and the subsequent drawing of the horizontal and vertical center axis. The intersections A₁ and A₃ thus result on the horizontal and the intersection B on the vertical center axis. Now the point A₄ is determined as desired at the circle K_R and then A₄ becomes connected with the points A₃, A₁ and B. The distance BA₄ is, according to the Südpolsatz, the angle bisector of the right angle A₁ A₄ A₃ in this case, since the point B lies on the perpendicular bisector of the triangle A₁ A₃ A₄. BA₄ intersects the distance A₁A₃ in O' with the distance d to the central point O of the circumcircle K_R. A parallel to the distance A₁A₄ follows from the point O'. The intersection point C produced here forms the distance O'C, the length of which is equal to the radius r, with which the inscribed circle K_r is now drawn around the point O'. The inscribed circle K_r contacted the distance A₁A₄ at the point D. Further, the two parallels to the vertical center axis proceed from the points D and C to the inscribed circle K_r. The intersections A'₁ and A'₃ are on the distance A₁O or E and F on the inscribed circle K_r. Finally, the two last sides of the quadrilateral A₁A₂ and A₂A₃ are drawn through the contact points E and F, respectively. Thus, the bicentric quadrilateral A₁ A₂ A₃ A₄ is completed.

comments

My biggest concern is, assuming i haven't misread the construction, that it does actually construct an arbitrary/general bicentric quadrilateral but "just" a bicentric (right angled) kite. The latter is proper construction in its own right, but it imho is not a construction that belongs in this section, which should give a general construction and not special cases. Aside from that there are some formal and language issues that need to be fixed:

• The article should not link to math theorem in de.wp or arguing based on them. Instead it needs to link to the appropriate article in en.wp or describe their needed content/math properties in English in this article.
• There are some issues concerning the use of overline as it creates a nonsensical notation in case of the indexed variables.
• distance (German: Abstand/Distanz) is falsely used as line segment (German: Strecke) at some places.
• if the construction indeed just yields a bicentric kite it probably can be somewhat shortened, the quote from Coolidge books for instance is if i'm not mistaken just a special case of Poncelet's porism.

--Kmhkmh (talk) 18:44, 2 August 2017 (UTC)

P.S. I categorized the drawing properly on commons now, so it is avaible via the commons link in the article as a special example/construction of a bicentric quadrilateral.--Kmhkmh (talk) 19:10, 2 August 2017 (UTC)

Improvement "Construction"

A bicentric quadrilateral ABCD with the contact quadrilateral WXYZ, animation see here

There is a simple method for constructing a bicentric quadrilateral.

It starts with the incircle C_r around the centre I with the radius r and then draw two to each other perpendicular chords WY and XZ in the incircle C_r. At the endpoints of the chords draw the tangents a, b, c and d to the incircle. These intersect at four points A, B, C and D, which are the vertices of a bicentric quadrilateral. It goes on with the two perpendicular bisectors p₁ and p₂ on the sides of the bicentric quadrilateral a respectively b. The perpendicular bisectors p₁ and p₂ intersect in the centre O of the circumcircle C_R with the distance x to the centre I of the incircle C_r. Finally, draw the circumcircle around the centre O.

The validity of this construction is due to the characterization that, in a tangential quadrilateral ABCD, the contact quadrilateral WXYZ has perpendicular diagonals if and only if the tangential quadrilateral is also cyclic.--Petrus3743 (talk) 16:03, 11 August 2017 (UTC)

Nice work on that new construction! I hope you didn't mind me including it into the main article. Circlesareround (talk) 06:11, 12 August 2017 (UTC)

@Circlesareround,

thanks, I'm glad if the improvement succeeded. It was true of Kmhkmh that he has requested an improvement. So an editing is fun! Greetings from Munich Petrus3743 (talk) 07:05, 12 August 2017 (UTC)

References

Incorrect Inscribed Circle Formula

The article makes reference to the MathWorld article for "r" and yet uses a wrong equation. The equation at MathWorld uses the semi-perimeter but this this article does not. — Preceding unsigned comment added by 81.187.174.44 (talk) 08:42, 20 October 2024 (UTC)