steiner point

梦里伊人 posted @ 2007年9月03日 04:53 in 其它 , 2421 阅读

Constructing a Steiner point of a Triangle



Assumption:  The Steiner point is an interior point of a triangle.  Any pair of rays extending from the Steiner point through any two vertices of a triangle form a 120 degree angle.


Construction Strategy:  Choose any two vertices of the triangle and consider the set of all intersections of rays emanating from each vertex where the angle of intersection of the rays is 120 degrees.  One of these must be the Steiner point.  Now consider the same set of intersections for a different pair of vertices.  This set must also contain the Steiner point.  Now find the unique intersection (excluding their common endpoint) of these two sets of intersections. This is the Steiner point.


How can the set of intersections described above be constructed?  We’ll start with the theorem in geometry stating that a vertex angle of a circle measures half the intercepted arc.  Consider a chord of a circle that intercepts 120/240 degrees of arc with endpoints A, B.  Choose any point on the 120 arc, C.  Angle ACB must be 120 degrees as it intercepts an arc of 240 degrees. 




Hence the set of points on the 120 arc describes the exact locus of points (intersections of rays) suggested above for two vertices of a triangle.  In this case, the vertices are represented by the endpoints of the chord. 


How do you construct a 120 arc pinned at the endpoints of a segment?  Let A,B be the endpoints of the segment, P be the center of the larger circle (not yet constructed), and M be the midpoint of AB.  Let’s assume the segment has length 2L.  Consider the whole sector containing this segment and the intercepted 120 degree arc. 


Triangle MPB is a 30-60-90 triangle and hence, if MB has length L, PB must have length .  PB is the radius of the circle.  Using the endpoints of the segment we could construct the center of the circle P if we could construct the length  knowing L.


How do you construct the length  (or ) knowing L?  First, construct a circle of radius L.  Then construct a second circle of radius L whose center lies on the original circle.  The two centers and one of the intersection points form an equilateral triangle of side length L.  Construct a perpendicular bisector of one of the angles of the triangle.  This divides the equilateral triangle into two 30-60-90 triangles with side lengths L, L/2 and .  Here is the diagram:





So we have a segment of length ; how do we get one of length ?


Draw a line and mark off the length  (four times the third length) and label the endpoints X,Y.  Now we just need to divide this segment into thirds.  Construct a line parallel to the original line.  Set the compass length to any length that is not this desired third.  Mark off 4 points G,H,I and J on the new parallel line equally spaced by this length.  Now draw lines through XG and YJ.   These lines should meet at a point, R.  Now draw lines through RH and RI:


These lines will divide our original segment XY into thirds.  We now have the desired length.



Back to the original problem:  Using the endpoints A and B, and the length we just constructed, mark off a point C a distance  from both A and B (by overlapping arcs).  This is the center of the circle of radius .  Using this center, construct the 1200 arc pinned at the ends of the segment AB. The Steiner point must lie on this arc.


Now go through the same process again, but this time starting on segment AC instead of AB. Here’s what we have:


The Steiner point must lie on the both of the 1200 degree arcs we have constructed, so it lies on their intersection; we have shown it in the above diagram. (If we had constructed a 1200 arc on the remaining side BC, it also would pass through this Steiner point.)




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