Urquhart theorem ("most elementary theorem of geometry")

vinoth_90_2004

Let ACE be a triangle, B on AC and D on AE. Let F be the intersection of CD and BE. Prove if AB+BF=AD+DF , then AC+CF=AE+EF.

**Posted:** Sat Jul 03, 2004 5:37 pm

darij grinberg

Indeed, AB + BF = AD + DF holds if and only if the quadrilateral ABFD has an excircle, and AC + CF = AE + EF holds if and only if the quadrilateral ACFE has an excircle. And both conditions are just equivalent to the existence of a circle touching the lines AC, AE, CF and EF.

The result is called Urquhart Theorem, for M. L. Urquhart (1902-1966). See

Dan Pedoe, The Most "Elementary" Theorem of Euclidean Geometry, Mathematics Magazine, 49 (1976) p. 40-42.

Darij

**Posted:** Sat Jul 03, 2004 10:30 pm

_________________
Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...

vinoth_90_2004

**Posted:** Sun Jul 04, 2004 4:38 am

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

darij was not quite exact saying "excircle", but we can always make it correct saying a circle that touches every side, probably with imaginary radius for your cases where two sides cross Mr. Green

(which is not important for the task anyway).

Peter

**Posted:** Sun Jul 04, 2004 5:21 am

darij grinberg

I knew my posting would entail a hell of confusion. Indeed, different forms of quadrilaterals and different kinds of in/excircles have different conditions of existence. When I was studying Mannheim's theorem, I had been trying to classify all types of quadrilaterals - convex, non-convex, self-intersecting - with all forms of tangent circles, but my patience was at the end very quickly. What I came up with is an application of the theory of directed circles providing a way of working with such quadrilaterals without having to distinguish dozens of cases.

We fix the orientation of a plane, i. e. we introduce a "clockwise" and a "counterclockwise" sense of rotation.

At first, if k is any circle in the plane, and d is a number from the set {-1; 1}, then we will call the pair (k; d) a directed circle. The radius of this directed circle is defined as the radius of the circle k, multiplied with d (so, the absolute value of the radius of the directed circle (k; d) equals the radius of k, and the sign is that of d).

Moreover, we consider directed lines. If g is a line in the plane, then there exist two unit vectors $\overrightarrow{e_1}$ , $\overrightarrow{e_2}$ parallel to this line. We have $\overrightarrow{e_1}=-\overrightarrow{e_2}$ . Now, if we pick out one of these two unit vectors - let's say, $\overrightarrow{e_1}$ -, then we can introduce directed segments on the line g by defining the directed distance PQ between two points P and Q on g to be the real number k such that $\overrightarrow{PQ} = k \cdot \overrightarrow{e_1}$ . It's clear that such a number exists and is unique. So we can call the pair (g; $\overrightarrow{e_1}$ ) a directed line; on any directed line, we can measure directed distances.

Now we are going to define the directed distance from a point to a directed line:

If P is a point, and (g; $\overrightarrow{e_1}$ ) is a directed line, then we consider the perpendicular p from the point P to the line g. This perpendicular p meets the line g at a point Q. If we now rotate the unit vector $\overrightarrow{e_1}$ about 90? in clockwise direction (i. e. about -90?, mathematically speaking), then the resulting vector $\overrightarrow{e_3}$ is parallel to p, and hence it defines, together with p, a directed line (p; $\overrightarrow{e_3}$ ). Now, on this directed line, we measure the directed distance PQ. This directed distance will be called the directed distance from the point P to the directed line g.

Finally, we will say that a directed circle is tangent to a directed line if and only if the directed distance of the center of the circle to the directed line equals the radius of the directed circle.

Actually, you will see that if a directed circle (k; d) is tangent to a directed line (g; $\overrightarrow{e_1}$ ), then the non-directed circle k must obviously touch the non-directed line g, but the converse is not true: If the non-directed circle k touches the non-directed line g, then the directed distance from the center of k to the directed line (g; $\overrightarrow{e_1}$ ) is either equal to the radius of the directed circle (k; d), or it has the same absolute value but different sign. In the first case, the directed circle is tangent to the directed line; in the second case, it is not.

In order not to lose the contact to earth in Bourbakist excesses, I will tell you how you can intuitively imagine the tangency of a directed circle with a directed line: You have a directed circle (k; d) and a directed line (g; $\overrightarrow{e_1}$ ). Consider a stone rotating uniformly and continuously along the circle. Namely, it should rotate counterclockwise if d = 1 and clockwise if d = -1. Suddendly, at the point where the circle k meets the line g, the stone flies away tangentially. Of course, it then flies along the line g. Now, if the direction in which the stone gets catapulted is equal to the direction of $\overrightarrow{e_1}$ , then the directed circle (k; d) is tangent to the directed line (g; $\overrightarrow{e_1}$ ).

Now we are ready to formulate the quadrilateral theorem:

If ( $g_1$ , $\overrightarrow{e_1}$ ), ( $g_2$ , $\overrightarrow{e_2}$ ), ( $g_3$ , $\overrightarrow{e_3}$ ), ( $g_4$ , $\overrightarrow{e_4}$ ) are four directed lines in the plane, then there exists a directed circle tangent to all of these four directed lines if and only if the equation $A_{12}A_{23}+A_{34}A_{41} = A_{23}A_{34}+A_{41}A_{12}$ holds, where $A_{ij}$ is the point of intersection of $g_i$ and $g_j$ (cyclic indices), and the distances $A_{41}A_{12}$ , etc. are directed distances measured along the directed lines ( $g_1$ , $\overrightarrow{e_1}$ ), ( $g_2$ , $\overrightarrow{e_2}$ ), ( $g_3$ , $\overrightarrow{e_3}$ ), ( $g_4$ , $\overrightarrow{e_4}$ ).

This is an arrangement-independent version of the well-known fact that a quadrilateral ABCD has an incircle if and only if AB + CD = BC + DA, and of all pendants of this fact related to excircles and concave quadrilaterals. You can also prove it in an arrangement-independent way (using the angle bisector between two directed lines - the set of all points having equal directed distances to these two lines).

Now you have the arsenal to prove Urquhart's theorem for all possible and impossible Mr. Green

cases at once. It is, BTW, a nice exercise on directed lines.

Concerning your comment, Peter, one doesn't need imaginary radii, but it's indeed impossible to distinguish between "incircles" and "excircles" in general.

Darij

**Posted:** Sun Jul 04, 2004 8:46 am