**Bi-centric Polygons**

Triangle is the basic polygon of 3 sides and 3 angles. Circle is considered as a polygon of infinite sides. In fact any closed figure can be considered as a polygon. Circle has a centre and all points on the circle are equidistant from the centre. Every regular polygon has a unique centre from where all sides are equidistant and all corners are also equidistant. In the case of triangles, the equilateral triangle has a unique centre as above. With this centre we can draw two circles one inscribed within the sides of the regular polygon and the other circumscribing the corners of the polygon. Such polygons which have two centres, a circum-centre and an in-centre are called bi-centric polygons. In case of regular polygons these two centres coincide. Irregular polygons can also be by-centric, but with displaced circum-centre and in-centre.

**Bi-Centric Triangles:**

For instance all triangles are inherently bi-centric. They have an in-circle tangential to all the three sides with an in-centre, I. This is always interior to the triangle. Triangles also have a circum-circle circumscribing the three corners of the triangle with a circum-centre, O. This can even be exterior to the triangle. In the case of equilateral triangle, as below, these two centres, I and O, coincide, with radius r of the in-circle, being half of radius R of circum-circle.

An interesting feature is: between these two circles you may inscribe infinite number of equilateral triangles, as shown in dotted lines, starting from any arbitrary point on the outer circle.

Now let us take another triangle, an isosceles triangle with its circum-circle and in-circle as shown below. This bi-centric triangle has its circum-centre and in-centre separated by a distance d.

Let OA = R, IE = r and OI = d.

∆ABD ||| ∆AEI. Hence IE/AI = BD/AD

i.e r/(R+d) = BD/2R ——– (1)

As, ∟CBD = ∟CAD = A/2

∟IBD = A/2 + B/2 = ∟BID

(i.e.) BD = ID = R-d

Hence from Eqn (1), r/(R+d) = (R-d)/2R

i.e. 2Rr = R^{2} – d^{2}

or, 1/(R+d) + 1/(R-d) = 1/r

This is a handy relationship between R, r and d. Now, like in the case of equilateral triangles, let us see whether these two circles also give rise to a family of triangles inscribed between them.

Let us take the same circles as above. Let us draw a tangent PQ to inner circle, from any point P on the outer circle. From P let us also draw a tangent PR to inner circle as shown, to a point R on the outer circle. Will QR be also a tangent to inner circle? Let us check. Draw PI and extend it to S on the outer circle.

With PQ and PR as tangents, PI is a bisector of angle P and hence

QS = RS and ∟SRQ = ∟SQR = P/2

Draw SO and extend the same to T on the outer circle. Join TQ.

∟SPQ = ∟STQ = P/2 = ∟SRQ = ∟SQR

Now,

PI.IS = AI.ID = (R+d).(R-d) = 2rR, (as derived earlier for these two circles)

i.e., PI/r = 2R/IS – – – – – – – (1)

But ∆PIE ||| ∆TSQ and hence,

PI/r = 2R/QS – – – – – – – – – – (2)

Comparing (1) and (2) above,

IS = QS = RS.

∟IRS = ∟RIS and ∟IQS = ∟QIS.

∟IRQ = ∟IRS – ∟SRQ = ∟RIS – P/2 = ∟IRP

Similarly ∟IQR = ∟IQP

i.e., IR and IQ are respectively the bisectors of angles PRQ and PQR.

Hence QR is also a tangent to the inner circle.

So, if you draw a tangent PQ to inner circle from any point P on the outer circle and continue drawing such tangents it will close on P again forming a triangle. Hence, like in the case of equilateral triangles, these two circles also give rise to a family of triangles inscribed between them. Amazingly this is true of all bi-centric polygons.

**Bi-Centric Polygons:**

Jean-Victor Poncelet (1788 – 1867) was a French mathematician who formulated this amazing feature in his Poncelet theorem. He found that this property is true even with polygons inscribed between two ellipses instead of circles as above. In one of its simpler forms, the Poncelet’s Closure Theorem (or Poncelet’s Porism) can be stated as below:

**If an n-sided polygon, n-gon, inscribed between two given ****conic sections****, is closed for one point of origin of the polygon, then it is closed for any position of the point of origin.**

Given two ellipses, one inside the other, if there exists an n-gon, simultaneously stays inscribed in the outer ellipse and circumscribes the inner ellipse, then any point on the boundary of the outer ellipse is a vertex of another such circum-inscribed n-gon of same number of sides. In the case of circles, their centres will become the circum-centre and in-centre of those n-gons and hence such n-gons are called Bi-centric Polygons.

*(Ref: **http://mathworld.wolfram.com/PonceletsPorism.html**)*

The following diagram shows the closure feature for ellipses:

* *

**Bi-Centric Quadrilaterals:**

Coming back to circles, we know of quadrilaterals which are circum-cyclic; i.e., its four vertices lie on a circle. We generally call them as cyclic quadrilateral. But not all quadrilaterals are circum-cyclic or circum-centric. For quadrilateral ABCD to be circum-centric, necessary and sufficient condition is: the opposite angles should add up to 180 degrees. i.e. Angles A+B = Angle C+D =180 Degrees. Same way, a quadrilateral can also be in-centric or in-cyclic. Necessary and sufficient condition for the same is: the opposite sides of the quadrilateral ‘abcd’ should add up to same value, i.e. a+c = b+d = s. Obviously those quadrilaterals which satisfy both the above conditions will be by-centric having both circum-centre and in-centre. The above conditions are easy to prove and understand. Looking at some of the standard quadrilaterals we can say:

Squares are all bi-centric. Rectangles are only circum-centric. Parallelograms can be only in-centric, as in case of Rhombus. Isosceles Trapeziums are circum-centric and can become bi-centric if opposite sides add up to same value. There are many other quadrilaterals which can qualify to be bi-centric. Another standard quadrilateral ABCD, known as a right-kite is also bi-centric as we will discuss below.

Let us consider the simplest example of a Bi-Centric quadrilateral ABCD, known as a right-kite, i.e., Angle ABC = Angle ADC = 90 degrees, AD = AB and CD = CB.

I is the centre of the in-circle of radius r. O is the centre of the circum-circle of radius R

‘d’ is the distance between them.

Angles ADC and ABC are 90 deg each and are lying in semi-circles.

Hence, Angles IAE + ICF = θ+ φ = 90 Deg.

Sin^{2}θ + Sin^{2}φ = Sin^{2}θ + Cos^{2}θ = 1

From ∆AIE, Sin^{2}θ = r^{2}/(R-d)^{2} and from ∆ICF, sin^{2}φ = r^{2}/(R+d)^{2}

Hence, r^{2}/(R-d)^{2 }+ r^{2}/(R+d)^{2} = 1

(i.e.) 1/r^{2} = 1/(R+d)^{2} + 1/(R- d)^{2}

This is again a very handy relationship between R, r and d.

We may write the above in a more interesting way as below

1/ID^{2} + 1/IB^{2 }= 1/(2r^{2}) +1/(2r^{2})

= 1/(R+d)^{2}+1/(R- d)^{2} = 1/IA^{2} + 1/IC^{2}

**For these two circles let us check the Poncelet’s theorem of closure**

Now let us construct an arbitrary quadrilateral PQRS inscribing the inner circle. Let PQ and QR be tangents to the inner circle, starting from any arbitrary point P on the outer circle. And let the tangents PS and RS meet at S. For Poncelet’s theorem to be true, S should lie on the outer circle. Draw PI and extend the same to E on the outer circle. Draw RI and extend the same to F on the outer circle. Let angle IPQ = θ and angle IRQ = φ.

∟ EOQ = 2θ and ∟ FOQ = 2φ

∟ EOQ + ∟ FOQ = 2 (θ +φ) – – – – – – — – – – – – (1)

Sin θ = r/PI = (r . IE) / (PI . IE) = (r . IE) / (AI . IC)

= (r . IE) / [(R+d).(R-d)]

i.e. Sin^{2} θ = (r^{2}.IE^{2}) / [(R+d)^{2}.(R-d)^{2}]

Now we know,

1/r^{2} = 1/(R+d)^{2} +1/(R-d)^{2} = 2(R^{2}+d^{2}) / [(R+d)^{2}.(R-d)^{2}]

Hence Sin^{2} θ = IE^{2}/[2(R^{2}+d^{2})]

|||ly, Sin^{2} φ = IF^{2}/[2(R^{2}+d^{2})]

So, Sin^{2}θ + Sin^{2}φ = (IE^{2} + IF^{2})/[2(R^{2}+d^{2})]

Now, IE^{2} = EO^{2} + IO^{2} – 2EO.IO.Cos(EOI)

= R^{2} + d^{2} – 2Rd Cos(EOI)

and IF^{2} = EO^{2} + IO^{2} – 2FO.IO.Cos(FOI)

= R^{2} + d^{2} – 2Rd Cos(FOI)

Hence, Sin^{2}θ + Sin^{2}φ = 1- K[Cos(EOI) + Cos(FOI)] – – – – (2)

Where K = [Rd/(R^{2} + d^{2} )]

∟ EOI + ∟ FOI = 360 – (∟ EOQ +∟ FOQ) = 360 – 2 (θ +φ)

Now, we know θ < 90 Deg and φ < 90 Deg, and (θ +φ) < 180 Deg.

Let us consider three cases in equation 2:

Case-1 : θ + φ < 90 Deg, and hence ∟EOI + ∟ FOI > 180 Deg

Sin^{2}θ + Sin^{2}φ < 1 and Cos(EOI) + Cos(FOI) < 0, RHS > 1

Case-2 : θ + φ > 90 Deg, and hence ∟ EOI + ∟ FOI < 180 Deg

Sin^{2}θ + Sin^{2}φ > 1 and Cos(EOI) + Cos(FOI) > 0, RHS < 1

Case-3 : θ + φ = 90 Deg, and hence ∟ EOI + ∟ FOI = 180 Deg

Sin^{2}θ + Sin^{2}φ = 1 and Cos(EOI) + Cos(FOI) = 0, RHS = 1

Hence only θ + φ = 90 Deg is admissible.

(i.e.) ∟ SPQ + ∟ SRQ = 180 Deg.

It follows from above, that quadrilateral PQRS is cyclic and hence S should fall on the same circle as PQR.

Thus for any arbitrary initial point P, PQRS will be a closed bi-centric quadrilateral, thus proving the Poncelet’s theorem for bi-centric quadrilaterals also.

**Conclusion:**

In short, let: a = 1/(R+d), b = 1/(R-d) and c = 1/r,

(where R= Radius of Circum-Circle, r = Radius of In-Circle of a bi-centric polygon and d= distance between them)

Then for bi-centric triangle, a+b = c,

for bi-centric quadrilateral, a^{2} + b^{2} = c^{2}, and

for bi-centric pentagon, a^{3 }+ b^{3 }+ c^{3} = (a+b)(b+c)(c+a)

or, (a + b + c)^{3} = 4(a^{3 }+ b^{3 }+ c^{3})

(ref: http://mathworld.wolfram.com/PonceletsPorism.html)

For a given R and d, the radius r of the inner circle will keep on increasing from (R^{2}-d^{2})/2R for a triangle, to (R^{2}-d^{2})/[√2√( R^{2}+d^{2})] for a quadrilateral asymptotically approaching R as number sides increases.

With these observations we will end this presentation of bi-centric polygons.

October 8, 2014 at 8:34 pm |

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