Let
XY denote the vector from point X to point Y.
Let P be a point in the plane of triangle ABC.
Let
PQ =
PA +
PB +
PC.
What is the
locus of points Q as P traces the triangle ABC ?
The program:
Dim x(3), y(3)
Private Sub cmdGo_Click()
Cls
x(1) = Val(txtX1): y(1) = Val(txtY1)
x(2) = Val(txtX2): y(2) = Val(txtY2)
x(3) = Val(txtX3): y(3) = Val(txtY3)
yScl = ScaleHeight / 10
Line (x(1) * yScl, ScaleHeight  y(1) * yScl)(x(2) * yScl, ScaleHeight  y(2) * yScl)
Line (x(2) * yScl, ScaleHeight  y(2) * yScl)(x(3) * yScl, ScaleHeight  y(3) * yScl)
Line (x(3) * yScl, ScaleHeight  y(3) * yScl)(x(1) * yScl, ScaleHeight  y(1) * yScl)
started = False
For pt1 = 1 To 3
pt2 = (pt1 Mod 3) + 1
For frac = 0 To 1 Step 1 / 1000
px1 = x(pt1) + frac * (x(pt2)  x(pt1))
py1 = y(pt1) + frac * (y(pt2)  y(pt1))
newx = x(1) + x(2) + x(3)  2 * px1
newy = y(1) + y(2) + y(3)  2 * py1
If started Then
Line (newx * yScl, ScaleHeight  newy * yScl)
Else
PSet (newx * yScl, ScaleHeight  newy * yScl)
started = True
End If
Next
Next
End Sub
shows for variously placed (via (x,y) values in text boxes on the VB form) vertices of triangle ABC that the resulting locus is a circumscribed triangle whose sides are parallel to the original triangle. Each side of the circumscribing triangle (the locus) goes through the vertex opposite the side of ABC that it is paralleling. The linear measures are twice those of the original triangle.
Justification:
Take a point P that is x units away from point A along AB. Let c be the length of side AB. The vector PQ then has added to it the resultant of PB and PA. This is certainly parallel to AB, and in magnitude it is x units toward A plus cx units toward B, for a net magnitude of c2x toward B. The fact that it is parallel to B makes all such points reside on a line parallel to AB. As x traverses a distance of c, the magnitude traverses 2c: when x is zero, the point Q is along the new line c units in the same direction as B from A; when x is c, Q is along the new line c units in the opposite direction. The same applies for all three sides of ABC, and the vertices of the original triangle, ABC, fall at the center of the circumscribing triangle.

Posted by Charlie
on 20080515 16:44:16 