A short history of the proof of an interesting
theorem
by Michal Stukow
An abbreviated version of the gold medal winning
paper at the Polish School Mathematical Contest in September 1996.
The abbreviations are due to the Editors.
We consider here tetragons which have both a circumscribed and
an inscribed circle. In the entire paper R represents the
radius of the circumscribed circle, whereas r is the radius
of the inscribed one. The centres of these circles are denoted
by O (the circumscribed) and I (the inscribed) and
w=OI is the distance between the centres. Moreover, S
is the tetragon's area and p is half its perimeter. We
aim at describing w in terms of R and r,
i.e. at finding a formula analogous to Euler's formula, valid
in every triangle:
1. Consider first a simple special case (see Fig. 1), when
the tetragon ABCD is a trapezium, i.e. has two sides parallel;
let their lengths be a and b with a greater
or equal to b. The trapezium has to be isosceles, for otherwise
no circumscribed circle
Fig. 1
would exist, and the length of the other two sides must be equal
to (a+b)/2, for otherwise there would be no inscribed circle.
Let K, L, M be the midpoints of the sides AB, BC, CD,
respectively. Then the centre I of the inscribed circle
is the midpoint of KM, whereas O lies on the straight
line MK in such a way that I belongs to OM.
By Pythagoras' theorem we have the following:
and since BL=IL=(a+b)/4, we obtain
Thus it suffices to determine
in terms of R and r. With the notation of Fig. 2
we infer from the sine formula the equality m=2Rsin.
Moreover,
Fig. 2
by Pythagoras' theorem
and sin= HD/AD =
4r/(a+b). By eliminating m and sinfrom
the three equations we arrive at
Now we can view this as a quadratic equation with variable k
and parameters R and r. After determining k
and applying simple transformations we get
(1)
2. The second special case will be that of a deltoid ABCD
with two opposite right angles (see Fig. 3).
Fig. 3
The point O coincides with the middle of the segment AC,
and I lies on the bisector of the angle ADC, so
by the properties of the bisector we now have
Thus AI=2aR/(a+b) and
It now remains to determine in
terms of the radii of the two circles. To do that let's denote
by S the area of the deltoid (it is equal to ab)
and let p=a+b be half of its perimeter.
Fig. 4.
The area S of an arbitrary polygon circumscribed
on a circle of radius r is equal to the product of half the perimeter
of the polygon by the radius of the circle.
Then S=pr, which further implies
,
the second equation following from Pythagoras' theorem applied
to the triangle ADC.
As before, we solve the quadratic equation with respect to p
and after some simple computation we arrive at formula (1)  the
same which holds for the isosceles trapezium.
The consistency of the results for two different cases suggests
the following conjecture:
The formula (1) holds for any tetragon
inscribed into a circle of radius R and circumscribed on a circle
of radius r.
3. Let's turn now to the general case (see Fig.5, which
provides the notation used).
Fig. 5
In the proof of the conjecture we will make use of the following
formulas:
(2)
(3)
The first equation in (2) is the socalled Brahmagupta formula,
valid for every tetragon inscribed in a circle (the second follows
from the fact that it is also circumscribed on a circle). The
formula (3) is a rather easy generalisation of the equality S=abc/4R,
valid for every triangle  you may wish to produce the corresponding
proof yourself.
Let's rewrite formula (1) in the form .
By the theorem on secants we infer that the lefthand side is
equal to ID IK. Thus it is sufficient to express this
product in terms of R and r. Here is a sketch of
the proof.
The angles ABK and ADK are equal, which implies
(check it!) that KBI is a rectangle triangle. Taking into
account the relations between the angles and the sides of the
triangles KBI and EBI, we get
.
Multiplying both sides of the above equation by ,
we arrive at
(4)
.
The second equation follows form the theorem of sines, the third
 from Ptolemy's theorem, well know to the readers of Delta.
Let m=ac+bd. Then applying (2) and (3) we get
(5)
.
Since p=a+c=b+d, we can use Brahmagupta's formula to get
the following equalities .
Hence and similarly,
. Applying the two latter
equalities we can transform (5) into
.
Now we can easily compute
and insert this value into (4). After eliminating surds in the
denominator we have a proof of our conjecture.
It may be worth noticing that the consideration of particular
cases did indeed help us in finding a proof for the validity of
the equality in the
general case. To begin with, we were able to formulate the conjecture
claiming that w depends only on R and r.
Thus we knew what we wanted to prove and this made possible an
application of the theorem on secants. Arguments of this kind,
which show the way from the formulation of a conjecture to a complete
proof, are quite rare in school literature. That certainly is
a pity, since they are very instructive indeed.
Finally, let's note an interesting conclusion that can be drawn
from (1). Since , we
have
and the equality only holds for the square, as can be easily verified.
