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Locus of Triangle (Posted on 2008-05-15) Difficulty: 3 of 5
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 ?

See The Solution Submitted by Bractals    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer for insight; then the reason | Comment 2 of 5 |

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 c-x units toward B, for a net magnitude of c-2x 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 2008-05-15 16:44:16
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