# 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 sin from 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 so-called 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 left-hand 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. 