Van Schooten's Locus Problem

According to [H. Dorrie, #47], Franciscus van Schooten (1615-1660), a Dutch mathematician, treated the following problem in his Exercitationes mathematica, published in 1657:

Two vertices of a rigid triangle in a plane slide along two arms of an angle of the plane; what locus does the third vertex describe?

J. Casey in his Analytic Geometry (1885) also ascribes the problem to van Schooten but with an earlier reference (Organica Conicorum DescriptioThursday, February 25, 2010 1:08:32 PM, 1646, c. 3, Ex. Math. IV.) and a different solution.

The applet below illustrates the proof from [H. Dorrie, #47]. Point A slides on the X-axis, point B on the Y-axis. The axes meet at the draggable origin O; the endpoints of the axes are also draggable. To see an outline of the curve traced by vertex C, check the box "Trace". If the box "Define triangle" is checked, points A and B move independently so that one can define the shape of triangle ABC. When the box is not checked, moving one of the points A, B causing the other to move too so that the distance between them is preserved.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

What if applet does not run?

Draw a circle L through A, B and O. Let M be the center of L. Line CM intersects L in two points P and Q. Since PQ is a diameter of L, ∠POQ = 90°. But more is true. Circle L has the same radius for all positions of AB because both AB is of ficed length and ∠AOB is fixed for a given position of the two lines. So we can imagine a fixed segment AB and a mobile point O with a fixed angle subtended by AB. For a fixed AB, point C describes a circular arc, which completes a circle when the angle at O becomes supplementary.

Think of circle L and points P and Q as being rigidly connected to triangle ABC. The arcs AP and AQ change their location but not their angular magnitude, meaning that P and Q remain on straight lines through O. This allows for shifting a view point. Two points P and Q slide along two perpendicular fixed lines that meet at O. Point C is fixed on line PQ. This is a situation described by the trammel of Archimedes. (So, in particular, since in the right triangle OPQ, OM is the median to the hypotenuse and hence traces a circle with center O.) Thus C describes an ellipse with axes along the orthogonal lines OP and OQ.

Hubert Shutrick made an observation that the center M of circle L too is rigidly connected to triangle ABC. The quadrilateral ACBD is thus rigid too and, since ∠AOM = ∠OAM, it rotates with the same angular velocity as OM but in the opposite direction. Thus, again, C traces an ellipse with center O and half axes OM + MC and OM - MC, check the page Dynamic construction of ellipse and other curves.

Hubert further suggested that a good way to draw an ellipse is to take a rod QP of length the sum of the required half axes and one OM with half its length attached with a pivot to the midpoint M of QP so it can fold like a jack knife. Let C be the point on QP such that CP is the length of the minor half axis. A fixed pivot at O allows OM to turn and P is constrained to move along the line that is chosen for the major axis. Then, C describes half of the ellipse while Q wanders up and down the minor axis. You get the other half if you can get Q to pass O. Teachers could get students to do it with bits of cardboard and drawing pins.


  1. H. Dorrie, 100 Great Problems of Elementary Mathematics: Their History and Solution, NewYork, Dover, 1989.
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