A Condition For A Circumscriptible Quadrilateral To Be Cyclic

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A Condition for a Circumscriptible Quadrilateral to be Cyclic
Article · January 2008
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b
Forum Geometricorum
Volume 8 (2008) 103–106. b b
FORUM GEOM
ISSN 1534-1178
A Condition for a Circumscriptible

Quadrilateral to be Cyclic
Mowaffaq Hajja
Abstract. We give a short proof of a characterization, given by M. Radić et al,

of convex quadrilaterals that admit both an incircle and a circumcircle.
A convex quadrilateral is said to be cyclic if it admits a circumcircle (i.e., a circle

that passes through the vertices); it is said to be circumscriptible if it admits an
incircle (i.e., a circle that touches the sides internally). A quadrilateral is bicentric
if it is both cyclic and circumscriptible. For basic properties of these quadrilaterals,
see [7, Chapter 10, pp. 146–170]. One of the two main theorems in [5], namely
Theorem 1 (p. 35), can be stated as follows:
Theorem. Let ABCD be a circumscriptible quadrilateral with diagonals AC and
BD of lengths u and v respectively. Let a, b, c, and d be the lengths of the tangents
from the vertices A, B, C, and D (see Figure 1). The quadrilateral ABCD is
cyclic if and only if uv = a+c
b+d .
B B
b b b b
c c
C C
c c
r
v a a
d d
u
D d a A D d a A
Figure 1 Figure 2
In this note, we give a proof that is much simpler than the one given in [5].
Our proof actually follows immediately from the three very simple lemmas below,
all under the same hypothesis of the Theorem. Lemma 1 appeared as a problem
in the M ONTHLY [6] and Lemma 2 appeared in the solution of a quickie in the
M AGAZINE [3], but we give proofs for the reader’s convenience. Lemma 3 uses
Lemma 2 and gives formulas for the lengths of the diagonals of a circumscriptible
quadrilateral counterpart to those for cyclic quadrilaterals as given in [1], [7, § 10.2,
p. 148], and other standard textbooks.
Publication Date: May 1, 2008. Communicating Editor: Paul Yiu.
The author would like to thank Yarmouk University for supporting this work and Mr. Esam
Darabseh for drawing the figures.
104 M. Hajja
Lemma 1. ABCD is cyclic if and only if ac = bd.

Proof. Let ABCD be any convex quadrilateral, not necessarily admitting an incir-
cle, and let its vertex angles be 2A, 2B, 2C, and 2D. Then A, B, C, and D are
acute, and A + B + C + D = 180◦ . We shall show that
ABCD is cyclic ⇔ tan A tan C = tan B tan D. (1)
If ABCD is cyclic, then A+C = B +D = 90◦ , and tan A tan C = tan B tan D,
each being equal to 1. Conversely, if ABCD is not cyclic, then one may assume
that A + C > 90◦ and B + D < 90◦ . From
tan A + tan C
0 > tan(A + C) =
1 − tan A tan C
and the fact that A and C are acute, we conclude that tan A tan C > 1. Similarly
tan B tan D < 1, and therefore tan A tan C 6= tan B tan D. This proves (1).
The result follows by applying (1) to the given quadrilateral, and using tan A =
r/a, etc., where r is the radius of the incircle (as shown in Figure 2).
Lemma 2. The radius r of the incircle is given by
bcd + acd + abd + abc
r2 = . (2)
a+b+c+d
Proof. Again, let the vertex angles of ABCD be 2A, 2B, 2C, and 2D, and let
α = tan A, β = tan B, γ = tan C, δ = tan D.
P P P
Let ε1 = α, ε2 = αβ, ε3 = αβγ, and ε4 = αβγδ be the elementary
symmetric polynomials in α, β, γ, and δ. By [4, § 125, p. 132], we have
ε1 − ε3
tan(A + B + C + D) = .
1 − ε2 + ε4
Since A + B + C + D = 180◦ , it follows that tan(A + B + C + D) = 0 and hence
ε1 = ε3 , i.e.,
r r r r r3 r3 r3 r3
+ + + = + + + ,
a b c d bcd acd abd abc
and (2) follows.
Lemma 3.
a+c b+d
u2 = ((a + c)(b + d) + 4bd), and v 2 = ((a + c)(b + d) + 4ac).
b+d a+c
Proof. Again, let the vertex angles of ABCD be 2A, 2B, 2C, and 2D. Then
1 − tan2 A a2 − r 2
cos 2A = =
1 + tan2 A a2 + r 2
2
a (a + b + c + d) − (bcd + acd + abd + abc)
= , by (2)
a2 (a + b + c + d) + (bcd + acd + abd + abc)
a2 (a + b + c + d) − (bcd + acd + abd + abc)
= .
(a + b)(a + c)(a + d))
A condition for a circumscriptible quadrilateral to be cyclic 105
Therefore
v 2 = (a + b)2 + (a + d)2 − 2(a + b)(a + d) cos 2A
a2 (a + b + c + d) − (bcd + acd + abd + abc)
= (a + b)2 + (a + d)2 − 2
a+c
b+d
= ((a + c)(b + d) + 4ac).
c+a
A similar formula holds for u.
Proof of the main theorem. Using Lemmas 1 and 3 we see that

ABCD is cyclic ⇐⇒ac = bd, by Lemma 1
⇐⇒(a + c)(b + d) + 4bd = (a + c)(b + d) + 4ac
u2 c+a 2

⇐⇒ = , by Lemma 3
v2 b+d
u c+a
⇐⇒ = ,
v b+d
as desired. This completes the proof of the main theorem.
Remarks. (1) As mentioned earlier, Theorem 1 is one of the two main theorems
in [5]. The other theorem is similar and deals with those quadrilaterals that admit
an excircle. Note that the terms chordal and tangential are used in that paper to
describe what we referred to as cyclic and circumscriptible quadrilaterals.
(2) Let A1 . . . An be circumscriptible n-gon and let B1 , . . . , Bn be the points
where the incircle touches the sides A1 A2 , . . . , An A1 . Let |Ai Bi | = ai for i =
1, . . . , n. Theorem 2 states that if n = 4, then the polygon is cyclic if and only if
a1 a3 = a2 a4 . One wonders whether a similar criterion holds for n > 4.
(3) It is proved in [2] that if a1 , . . . , an are any positive numbers, then there
exists a unique circumscriptible n-gon A1 . . . An such that the points B1 , . . . , Bn
where the incircle touches the sides A1 A2 , . . . , An A1 have the property |Ai Bi | =
ai for i = 1, . . . , n. Thus one can, in principle, express all the elements of the
circumscriptible polygon in terms of the parameters a1 , . . . , an . Instances of this,
when n = 4, are found in Lemms 2 and 3 where the inradius r and the lengths of
the diagonals are so expressed. When n > 4, one can prove that r 2 is the unique
positive zero of the polynomial
σn−1 − r 2 σn−3 + r 4 σn−5 − . . . · · · = 0,
where σ1 , . . . , σn are the elementary symmetric polynomials in a1 , . . . , an , and
where a1 , . . . , an are as given in Remark 2. This is obtained in the same way we
obtained (2) using the the formula
ε1 − ε3 + ε5 − . . .
tan(A1 + · · · + An ) = ,
1 − ε2 + ε4 − . . .
where ε1 , . . . , εn are the elementary symmetric polynomials in tan A1 , . . . , tan An ,
and where A1 , . . . , An are half the vertex angles of the polygon.
106 M. Hajja
References
[1] C. Alsina and R. B. Nelson, On the diagonals of a cyclic quadrilateral, Forum Geom., 7 (2007)
147–149.
[2] D. E. Gurarie and R. Holzsager, Problem 10303, Amer. Math. Monthy, 100 (1993) 401; solution,
ibid., 101 (1994) 1019–1020.
[3] J. P. Hoyt, Quickie Q 694, Math. Mag., 57 (1984) 239; solution, ibid., 57 (1984) 242.
[4] S. L. Loney, Plane Trigonometry, S. Chand & Company Ltd, New Delhi, 1996.
[5] M. Radić, Z. Kaliman, and V. Kadum, A condition that a tangential quadrilateral is also a chordal
one, Math. Commun., 12 (2007) 33–52.
[6] A. Sinefakupoulos, Problem 10804, Amer. Math. Monthy, 107 (2000) 462; solution, ibid., 108
(2001) 378.
[7] P. Yiu, Euclidean Geometry, Florida Atlantic Univesity Lecture Notes, 1998, available at
http://www.math.fau.edu/Yiu/Geometry.html.
Mowaffaq Hajja: Mathematics Department, Yarmouk University, Irbid, Jordan

E-mail address: mhajja@yu.edu.jo, mowhajja@yahoo.com
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See discussions, stats, and author profiles for this publication at: https://www.researchgate.

A Condition for a Circumscriptible Quadrilateral to be Cyclic

Article · January 2008

Geometry View project

Ring Theory and Geometry View project

The user has requested enhancement of the downloaded file.

A Condition for a Circumscriptible

Abstract. We give a short proof of a characterization, given by M. Radić et al,

A convex quadrilateral is said to be cyclic if it admits a circumcircle (i.e., a circle

Lemma 1. ABCD is cyclic if and only if ac = bd.

Proof of the main theorem. Using Lemmas 1 and 3 we see that

Mowaffaq Hajja: Mathematics Department, Yarmouk University, Irbid, Jordan

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