Sketchpad Solutions and Results

Michael de Villiers, Dynamic Math Learning

In the July 1998 issue, a copy of the software package *Sketchpad* was offered as a prize for the most elegant set of solutions for the three investigations given below. Correct, elegant solutions were received from Jacques Conradie (Graad 9, Hoėrskool De Kuilen), Bruce Merry (Grade 10, Westerford High) and Neil Olver (Grade 11, Bergvliet High). The prize was, however, awarded to Neil who gave the most complete solution by looking at some of the special cases below. In investigations of this kind it is often useful (although not essential) to first use construction and measurement. However, paper and pencil construction and measurement is tedious and relatively inaccurate, whereas computer exploration by means of a programme like *Sketchpad* is quick and easy, and very accurate.

1. *Investigate the type of quadrilateral formed by reflecting the circumcenter of a cyclic quadrilateral in each of its four sides.*

Let the circumcentre of the cyclic quad ABCD be O, and its reflections in AB, BC, CD and DA be O1, O2, O3 and O4 respectively. Let the intersection of the lines OOi with the respective sides of ABCD be Ki. Since O1 is the reflection of O in AB, angle OK1B = 90°. Thus K1 is the midpoint of AB. Likewise K2, K3 and K4 are the midpoints of their respective sides.

But it is a well-known fact that quad K1K2K3K4 is a parallelogram (called the Varignon parallelogram). Since Ki is the midpoint of OOi by construction, it follows that O1O2O3O4 is simply an enlargement of K1K2K3K4 from centre O with a scale factor of 2. Therefore O1O2O3O4 is a parallelogram. (Note that a central similarity such as this is called *homothetic*, which means that the corresponding sides of the two similar polygons are always parallel to each other. This can easily be seen by looking for example at the corresponding linesegments K1K2 and O1O2 in the similar triangles OK1K2 and OO1O2.)

Note that the result and the above proof is also valid when ABCD is crossed. The above proof also immediately leads to the following generalization:

*If any arbitrary point is point-reflected through the midpoints of the sides of any quad ABCD, then the formed quadrilateral is a parallelogram (homothetic to the Varignon parallelogram with a scale factor of 2). *

[A point-reflection occurs when a point A is reflected through a point B to its image A' by extending the line AB beyond B to A' so that BA' = BA.]

2. *Investigate the type of quadrilateral formed by reflecting the incenter of a circum quadrilateral (a quad circumscribed around a circle) in each of its four sides.*

Let the incentre of the circum quad ABCD be I, and its reflections in AB, BC, CD and DA be I1, I2, I3 and I4 respectively. Let the intersection of the lines IIi with the respective sides of ABCD be Ki. Since angle IK1A = 90°, K1 must be the point of tangency between the circle and AB. Likewise for K2, K3 and K4.

But IK1 = IK2 = IK3 = IK4. Therefore II1 = II2 = II3 = II4 which implies that I1I2I3I4 is cyclic. Moreover, as in Question 1, I1I2I3I4 is homothetic to K1K2K3K4 with a scale factor of 2.

Note that the result and proof is also valid when the circum quad ABCD is concave or crossed as shown above. [In the concave case shown, BC and DC extended are tangent to the circle. In the crossed case shown, DA and BC extended are tangent to the circle. In both cases, the two homothetic cyclic quads are crossed.].

3. *Investigate the type of quadrilateral formed by reflecting the point of intersection of the diagonals of a cyclic quadrilateral in each of its sides.*

Let the intersection of the diagonals of cyclic quad ABCD be G, and its reflections in AB, BC, CD and DA be G1, G2, G3 and G4 respectively. Let the intersection of the lines GGi with the respective sides of ABCD be Ki. As before, G1G2G3G4 is homothetic to K1K2K3K4 with a scale factor of 2.

AK1GK4 and K4GK3D are both cyclic (angles AK1G, AK4G, DK4G and DK3G are all right angles). Therefore angle K1AG = angle K1K4G on chord K1G, and angle GK4K3 = angle GDK3 on chord GK3. But angle K1AG = angle GDK3 on chord BC. Therefore, angle K1K4G = GK4K3; ie. G lies on the angle bisector of angle K1K4K3. Similarly, G lies on the angle bisectors of the other three angles of K1K2K3K4; in other words, its angle bisectors are concurrent at G. Therefore K1K2K3K4 is a circum quadrilateral, and thus also G1G2G3G4.

It should be noted that when the circumcenter of convex quad ABCD falls on one of its sides, G1G2G3G4 degenerates into a triangle, and when it falls outside ABCD, G1G2G3G4 becomes concave (see first figure below). The result and proof (with some modifications) is also valid when ABCD is crossed, with G1G2G3G4 being convex, degenerate, concave or crossed (for the latter, see second figure below).

For ordering information about *Sketchpad*, and specialized mathematical publications, please contact Michael de Villiers at Tel: 031-2044252(w); 031-7083709(h); fax: 031-2044866(w); e-mail: dynamiclearn@mweb.co.za OR Pearl de Villiers at Tel: 031-4507305(w). Dynamic Learning, 8 Cameron Rd, Sarnia (Pinetown) 3615.