1 The Elusive Poncelet Point Revised

by Jae Hyun (James) Yang

Abstract

A theorem to express the coordinates of P with the coordinates of the initial triangle

was proven by Eugene Lee, and has shown significantly accurate and compelling results.

However, the theorem only considers a specific case - when the initial triangle is isosceles.

We now raise a natural question: can the theorem be generalized to any triangle? Although

Eugene’s theorem may work, the computation in the recursion process for a general triangle,

as Eugene states in the conclusion, ”seems like a daunting task” (18). Hence, we introduce a

different method using complex number system to express the coordinates of P in a simpler

and faster way.

By Eugene’s theorem, the y-coordinate is entirely expressed in terms of the vertices of the

initial triangle and an infinite sum of a trigonometric function. Because the infinite sum does

not seem to converge to a closed form, practically, obtaining the convergence point seems

highly demanding even with computers. This theorem is therefore driven by practicality,

not ideality. We will define a converging sequence that allows us to estimate the location of

the Poncelet point of any initial triangle with a minor recursion process.

1

Introduction

We have already established that for a given triangle T1, one may construct an infinite

sequence of contact triangles, which converges to a point (Poncelet point). However, many

interpretations arise regarding the converging sequence. One could interpret that the con-

tact triangles converge to P, which was the direction in Eugene’s theorem, and another could

interpret the inscribed circles converge to P. In this theorem, we will look at neither of those

sequences. Instead, we study the sequence of the inscribed circles’ center points. Already,

we have simplified the problem in that now we study the behavior of single points, rather

than that of a collection of points (triangle or inscribed circle), converging to P.

Figure 1: Complex plane representation of triangle

2

Figure 2: Cartesian plane representation of triangle

The purpose of introducing complex number system is to provide extra algebraic prop-

erties, such as the unification of sine and cosine with exponential function, all the while

preserving the shapes of figures. For example, in Fig. 1 and Fig. 2, the two triangles are

identical, but Fig. 1 is expressed in C-space, and Fig. 2 is expressed in R2-space. We can

express the vertices with coordinates (x,y), where x,y are real numbers. The coordinates de-

scribe a point’s location on both real and complex space, however the fundamental difference

is that the point in R2-space is simply a tuple of two real numbers, but in C, the tuple of

two real numbers represents another number (complex number), which has many algebraic

properties the point in R2 does not have.

3

Theorem

Lemma 1. Let z1, z2, and w be complex numbers such that w is along the line connecting z1

and z2. Then,

w = z1 + |w − z1| eiArg(z2−z1)

Proof.

w =z1 + (w − z1)=z1 + |w − z1| eiArg(w−z1)

=z1 + |w − z1| eiArg(z2−z1),

since w is on the line connecting z1 and z2, Arg (w − z1) = Arg (z2 − z1). This completes

the lemma.

Theorem 2. Let z1,n, z2,n, z3,n be complex numbers that represent the vertices, and An =

|z3,n − z2,n| , Bn = |z1,n − z3,n| , Cn = |z2,n − z1,n| be the lengths of the sides of the nth contact

triangle 4z1,nz2,nz3,n, where n = 0 represents the initial triangle 4z1,0z2,0z3,0. Let P be the

Poncelet point of z1,0z2,0z3,0 and (Pn), for n=0,1,2,..., be the converging sequence of center

points of the inscribed circles, where

Pn = z1,0 +n−1∑j=0

bjeiθj + bn sec

(αn2

)ei(θn−

αn2 ) (1)

and

bj =Bj + Cj − Aj

2(2)

θj = Arg (z2,j − z1,j) (3)

αn = Arg

(z2,n − z1,nz3,n − z1,n

)(4)

Then

P = limn→∞

Pn = z1,0 + limn→∞

n∑j=0

bjeiθj (5)

Proof. By definition, 4z1,1z2,1z3,1 is the first contact triangle, and P0 represents the center of

the first inscribed circle. Our first goal is to express P0, and ultimately, Pn, for n = 0,1,2...,

relative to 4z1,0z2,0z3,0.

4

Figure 3: Initial triangle with first contact triangle and inscribed circle

Let A0, B0, C0 be the lengths of the initial triangle and a0, b0, c0 be the sub-lengths of

A0, B0, C0 as shown in Fig. 3. Since lengths of tangents from any external point to a circle

are equal, we can subdivide the lengths A0, B0, C0 as a sum of a0, b0, c0 as shown in Fig. 3.

A0 = a0 + c0, B0 = b0 + c0, C0 = a0 + b0

By addition,

A0 +B0 + C0 = (a0 + c0) + (b0 + c0) + (a0 + b0)

=2 (a0 + b0 + c0)

=2 (a0 + c0 + b0)

=2 (A0 + b0)

A0 +B0 + C0

2=A0 + b0

b0 =B0 + C0 − A0

2

5

Similarly, we can express all sub-lengths in terms of the lengths of a given triangle. We can

generalize bn such that

bn :=Bn + Cn − An

2

By Lemma 1, the vertices of the nth contact triangle are formulated as such,

z1,n =z1,n−1 + bn−1eiArg(z2,n−1−z1,n−1)

z2,n =z2,n−1 + an−1eiArg(z3,n−1−z2,n−1)

z3,n =z1,n−1 + cn−1eiArg(z1,n−1−z3,n−1)

Notice that through recursion, zi,n is completely expressed in terms of zi,0, for i = 1, 2, 3.

From Fig. 3, we can express the sides of the initial triangle as complex numbers as such

z2,0 − z1,0 =Ceiθ1

z3,0 − z1,0 =Beiθ2

Dividing,

z2,0 − z1,0z3,0 − z1,0

=C

Bei(θ1−θ2)

Let,

α0 := θ1 − θ2 = Arg

(z2,0 − z1,0z3,0 − z1,0

)We can generalize αn such that

αn := Arg

(z2,n − z1,nz3,n − z1,n

)

6

Figure 4: Bottom left segment of Fig. 3

Considering Fig. 4, by side-angle-side congruency, 4z1,0z1,1P0∼= 4z1,0z3,1P0 and so,

−−−→z1,0P0 bisects α0.

r0 =∣∣∣b0 tan

(α0

2

)∣∣∣ = b0 tan(α0

2

)since r0 ≥ 0, b0 ≥ 0, and 0 ≤ α0 < π.

Since−−−→z1,1P0 is orthogonal to −−−−→z1,0z1,1, and therefore, to −−−−→z1,0z2,0 (on the same line), and is

turned 90o clockwise,

Arg (P0 − z1,1) = Arg (z1,1 − z1,0)−π

2

= Arg (z2,0 − z1,0)−π

2

7

Furthermore, letting θ0 = Arg (z2,0 − z1,0) = Arg (z1,1 − z1,0),

P0 − z1,0 = (z1,1 − z1,0) + (P0 − z1,1)

=b0eiθ0 + r0e

i(θ0−π2 )

=b0eiθ0 − ir0eiθ0

= (b0 − ir0) eiθ0

Substituting r,

=(b0 − ib0 tan

(α0

2

))eiθ0

=b0

(1− i tan

(α0

2

))eiθ0

=b0 sec(α0

2

)(cos(α0

2

)− i sin

(α0

2

))eiθ0

=b0 sec(α0

2

)e−i

α02 eiθ0

=b0 sec(α0

2

)ei(θ0−

α02 )

Therefore,

P0 = z1,0 + b0 sec(α0

2

)ei(θ0−

α02 )

We can generalize Pn such that

Pn := z1,n + bn sec(αn

2

)ei(θn−

αn2 )

where,

θn := Arg (z2,n − z1,n) = Arg (z1,n+1 − z1,n)

Let us consider the equation

z1,n = z1,n−1 + bn−1eiArg(z2,n−1−z1,n−1)

Then

z1,n − z1,n−1 =bn−1eiArg(z2,n−1−z1,n−1)

z1,n−1 − z1,n−2 =bn−2eiArg(z2,n−2−z1,n−2)

.

.

.

z1,3 − z1,2 =b2eiArg(z2,2−z1,2)

z1,2 − z1,1 =b1eiArg(z2,1−z1,1)

z1,1 − z1,0 =b0eiArg(z2,0−z1,0)

8

By telescoping method,

z1,n − z1,0 =n−1∑j=0

bjeiArg(z2,j−z1,j)

z1,n =z1,0 +n−1∑j=0

bjeiArg(z2,j−z1,j)

Hence,

z1,n =z1,0 +n−1∑j=0

bjeiθj

Therefore by substitution,

Pn = z1,0 +n−1∑j=0

bjeiθj + bn sec

(αn2

)ei(θn−

αn2 )

Since the sequence (Pn) converges to the Poncelet point P ,

P = limn→∞

Pn = z1,0 + limn→∞

n−1∑j=0

bjeiθj + lim

n→∞bn sec

(αn2

)ei(θn−

αn2 )

We know αn converges to π3

(more will be explained in Section 1) as n → ∞, and eiθn is

bounded. Since bn is strictly less than bi,∀i = 0, 1, . . . , n− 1 and the last term is bounded,

the last term converges to 0 faster than every other terms, so

P = z1,0 + limn→∞

n∑j=0

bjeiθj

where,

bj =Bj + Cj − Aj

2

θj = Arg (z2,j − z1,j)

Inherently, bj and θj are completely expressed in terms b0 and θ0, which are completely

expressed in terms of initial triangle vertices z1,0, z2,0, z3,0 through recursion, and therefore,

P is completely expressed in terms of z1,0, z2,0, z3,0. This concludes the proof.

9

Computational Accuracy

As n→∞, the computation for the Poncelet point P is incredibly daunting still. How-

ever, there is evidence for fast convergence of this sequence. By simple geometry, we see

that αnβnγn

=

0 0.5 0.5

0.5 0 0.5

0.5 0.5 0

αn−1βn−1

γn−1

which by induction gives thatαnβn

γn

=

0 0.5 0.5

0.5 0 0.5

0.5 0.5 0

n α0

β0

γ0

Let

A =

0 0.5 0.5

0.5 0 0.5

0.5 0.5 0

Since A is symmetric, it is diagonalizable with an orthonormal eigenbasis with all real eigen-

values by spectral theorem. Let C be an orthogonal matrix with column vectors as or-

thonormal eigenvectors in elementary basis and Λ be the diagonal matrix of eigenvalues.

Then,

A = CΛCT

where,

C =

1√3

1√2

1√6

1√3− 1√

21√6

1√3

0 − 2√6

,Λ =

1 0 0

0 −12

0

0 0 −12

An =

(CΛCT

)n=(CΛCTCΛCT . . . CΛCT

)= CΛnCT

limn→∞

An = limn→∞

C

1 0 0

0(−1

2

)n0

0 0(−1

2

)nCT = C

1 0 0

0 0 0

0 0 0

CT =

13

13

13

13

13

13

13

13

13

Therefore, αnβn

γn

=

0 0.5 0.5

0.5 0 0.5

0.5 0.5 0

n α0

β0

γ0

→

13

13

13

13

13

13

13

13

13

α0

β0

γ0

=

π3π3π3

10

This convergence is incredibly fast. A simple programming test shows that when n equals

seven, matrix A is already similar to the convergent matrix of 13.

We can easily show that for an initial equilateral triangle, the Poncelet point is the center

point of all inscribed circles. Therefore, the fast convergence of the angles implies a fast

convergence of the center points Pn.

Results

Figure 5: a) From top right to bottom left, sequence of circle center points. b) magnitude

difference between consecutive circle center points

Here is an example for a randomly generated triangle. Fig. 5 a) shows the convergence of

the center points Pn, for n = 0, 1, . . . , 29, on the complex plane. Notice the extremely linear

movement and the fast convergence of the points. Fig. 5 b) shows the consecutive distances

from one center point to another starting from the top right. Notice that the distances are

decaying exponentially.

11

Figure 6: Tenth center point and convergent point.

Fig. 6 shows where the tenth center point P9 and the Poncelet point P lie on the complex

plane. The absolute distance is 1.47798032012× 10−7, which is incredibly accurate for only

10 recursions.

Figure 7: Relative error graph

12

In Fig. 7, we show the relative errors of the estimate using P9. The error is calculated as

such

Error =|P − P9|b0 tan

(α0

2

)where b0 tan

(α0

2

)= radius of the first circle. The root mean square (RMS) of the error for

100 randomly generated triangles was 1.31427×10−6. A test was also performed to compare

the accuracy of P9 with Eugene’s formula for an isosceles triangle.

P9 =(−1.95156391047× 10−16, 1.70936028984

)Eugene = (0, 1.70936107043)

The x-coordinate can simply be assumed to be 0. The y-coordinates are exactly equal up to

the millionth place.

13