The Brocard Point - John Sharp
Here, the challenge is to find, for any triangle, the point which is the intersection of three lines equally inclined to the sides. This special point is known as the Brocard point.
This is not only a suitable subject for Dynamic Geometry packages, showing new insights into an old topic, but is also a good exercise in introducing techniques for the use of the packages. The following illustrations use Geometer’s Sketchpad. The Brocard points are usually written about in “advanced&rdsquo; books on Euclidean Geometry, whereas with this approach they can be easily understood at an average school level. Linked with the Miquel point theorem, they can show the beauty of mathematics.
First create your triangle from three points, A, B and C. We are looking for three lines which go though the vertices of the triangle and which are inclined at an equal angle to each of the side of the triangle. At one particular angle these three lines all intersect in the same point. The easiest way to find this point using Geometer’s Sketchpad is to create three lines and rotate them.
Now create a varying angle. This can be used to animate the rotation (which we will use later), but we need to work from one angle so that we ensure all rotations are the same.
- Draw a circle anywhere on the screen. This automatically creates two points, the centre of the circle D and a point on it which defines the radius, E.
- Create another point on the circle, F. Together with points E and F, this point defines an angle. It helps to see the angle if we create the lines DE and EF as well.
- To define the angle, select points E, then D then F while holding down the shift key.
- On the Transform menu, select Mark Angle EDF. Note how it is shown as an anti-clockwise movement.
Figure 1
We now need to create the three rotating lines. On each side of the triangle, create an arbitrary point, G on AB, H on BC and I on CA. We could use the vertices of the triangle, but it helps to avoid confusion if we work with other points. Using these points, create the lines as follows:
- Mark A as the centre of rotation by double clicking it. Then select point G and go to the Transform menu and choose Rotate. When the dialogue box comes up, you should see the By Marked Angle box ticked and you should also see By Angle D-E-F at the top. Click on the OK button to create the rotated point G'. Now create a line by joining A and G, making sure the line type is set to draw a complete line, with the type set as in Figure 1
- It is a good idea to colour each of the rotating lines differently.
- In the same way mark B and rotate point H and then create line BH', and then rotate I about point C and draw line CI'.
- From now on, it is easier to see what is happening if you hide points G', H' and I'.
Now if you rotate point F on the circle, you can adjust the angle until the three lines intersect in the same point. To see the value of the angle, display its measurement, and see how it varies as you alter the angles of the triangle. What values can it have as the triangle changes? Is there a maximum value, and if so, is the triangle special in any way?
Figure 2
It is not easy to make them coincide exactly, but by changing the angle, you can see that they do. There is a better way to find the Brocard point, by animating the intersections which leads to an accurate way to find the Brocard angle.
Tracing the points of intersection
By tracing the points of intersection of the three lines , the Brocard point becomes the common point for the three traced curves. This also helps to show that there is a single point of intersection and not just a small triangle.
- Select the lines AG', BH' and CI' in pairs and go to the Construct menu and mark the Point at Intersection.
- With each of these constructed points selected, go to the Display menu and make them points that are traced.
- Then to create an animation, select the circle, and the point F and on the Edit menu create an Animate action button.
When you animate the rotation, the three points of intersection move on three circles. These circles intersect in one point, the Brocard point.
Figure 3
This is still not accurate enough to find the Brocard point, and does not help if you want to determine how the Brocard angle varies as the triangle changes shape. To do this we have to look at an associated theorem for which the Brocard point is a special case.
The Miquel point
The mathematician Miquel discovered the following theorem in 1832. If a point is picked at random on each side of a triangle ABC, then the three circles constructed through each vertex and the points on the adjacent sides are concurrent.
Miquel’s theorem is easy to demonstrate in Geometer’s Sketchpad. You could create a new sketch, but you might like to use the one we have already constructed.
- If you are using the same sketch, it helps to hide the lines and traced points but to be able to show them later. Select the lines AG', BH' and CI' and their points of intersection and go to the Edit menu and create a pair of Hide/Show action buttons.
- If you are creating a new sketch, draw the triangle and place three points G, H and I on the sides as before.
- Select points A, G and I and go to the Construct menu and choose Arc Through Three Points. Do the same for points B H and G and the points C, I and H.
- Move the points along the lines to see that the three arcs always intersect in a single point. This is Miquel’s theorem.
This is an example of a Miquel point:
Figure 4
Now what has this to do with the case of the Brocard point? Well, it is obviously a special case. Before continuing, you might like to vary the position of points G H and I to match the Brocard circles case.
The answer is that for the three Miquel circles to give the Brocard point they must correspond to the degenerate case when G moves to A, H moves to B and I moves to C. You might also notice that this means that the circle through A and B has the line BC tangent at point B, that through A and C has tangent AB at A and that through B and C has tangent AC at C.
You should also be aware of is that there are two Brocard points for a triangle. The other one (in Miquel terms) occurs when G moves to B, H to C and I to point A. This is because the movement can be clockwise or anti-clockwise. The Brocard angle is the same whichever way you move.
The accurate Brocard points
Figure 5
Now we know that the Brocard points are formed by the intersection of three circles which are tangent at the vertices, instead of drawing the circles as a traced point, they can be drawn using the circle option. I leave that exercise to the reader to obtain a diagram like the following:
Special cases of Brocard points
Figure 6
Brocard arrived at his discovery of the points of a triangle named after him from the study of the problem of the three dogs chasing one another. This is a common curve stitching example, which is usually created using an equilateral triangle (or other regular polygon) where the three dogs all move at the same speed. Brocard was interested in the general case, but where the dogs still meet at the same time at the same point – the Brocard point. This means that they have to run at different speeds. Determining the speeds is more complicated and outside of the scope of the simple discussion here, but the following is an example of the type of curve stitching effect you could achieve.
Figure 7
When Brocard was working on this problem, at the end of the last century, the Euclidean geometry of the triangle was being studied intensely by many mathematicians. Hundreds of results were found relating to the Brocard point with a bibliography of nine pages in 1886 which was supplemented by another one in 1890 which covered 25 pages. Whole books were written on the geometry of the triangle including one devoted just to the Brocard point. So you might expect to see in many places. Figure 6 is one.
Figure 7 is a triangle whose sides are in the Golden section with angles 36 and 72 degrees. A similar triangle has been cut off and the same process repeated to give a kind of fractal effect. In the limit the triangle goes to a point at the Brocard point of the triangle.
Can you find any more instances of the Brocard point?
