Hamilton Hardison’s Exploration of

Assignment 6:  Locus of the Vertex of a Fixed Angle Subtended by a Fixed Segment

From Jim Wilson’s Assignments:  “Of a triangle, given two vertices A and B, and the angle at the third vertex C (the angle opposite side AB). What is the locus of the point C? See Script # 18 in Assignment 5.

(Note: Think of this as turning a fixed angle so that its sides always rest on the two endpoints of the segment AB. Remember that the size of the angle is fixed but the sides can move along A and B simultaneously.)”


By exploration (with or without technology), we may arrive at a picture like the following:

Such an exploration leads us to believe that the locus of the vertex point, C, will be the arc of a circle.  Using GSP we can develop a more elegant and convincing locus.  If we start with a segment AB and an angle at the third vertex C as the assignment suggests, we can create an arbitrary point, Drag, on the circle whose center is A and radius is AB.  This creates an angle .

Since the sum of the interior angles of any triangle is 180o, (or a straight angle), we can extend the initial side of the given angle whose vertex is C to a line.  Now we can add angle  and the angle whose vertex is C to discover the third angle of the triangle, .

We can now construct angle , using vertex B and side AB.

The intersection of rays emanating from A and B not containing the segment AB is one possible position of our vertex point C.  By using the locus capabilities of GSP, we can construct the path of C as the point Drag is moved.

It appears that our locus is an arc of a circle.  If this is the case, we should be able to find the center of the circle.  Since a circle is the set of all points equidistant from a fixed point, O, we must have that O lies on the perpendicular bisector of AB.  (This perpendicular bisector is the set of all points equidistant from A and B).  For construction purposes, we can build  in a special way so that it is 90o, (any acute angle will work, but 90o is easiest to construct).  The angle  is now the complement of the fixed angle whose vertex is C.  The complement of the angle whose vertex is C can now be constructed at point B using AB as the initial side.  The center, O, is the intersection of the perpendicular bisector of AB and the terminal side of angle .  (Similar triangles give that this intersection lies on the perpendicular bisector of segment AC’)

By exploration, we discover that the locus contains the minor arc of the circle when the angle whose vertex is C is obtuse.  A semicircle results when C is a right angle.


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