



 
Hi, Fix two vertices A and B to form one side of a triangle. Now choose another point C not on the line joining A and B and construct the triangle ABC. I can then construct a triangle ABD with a smaller perimeter than your triangle by placing D half way between A and C. The point here is that no matter what triangle you choose with side AB I can always construct a triangle with side AB that has a smaller perimeter than yours. Thus it's not possible to choose a third point that results in the smallest possible perimeter. Penny RS wrote back Thanks very much. Perhaps I did not make phrase my question correctly. With two points fixed and the third point variable but along a straight line, then which is the smallest triangle. Please see attached diagram. Regards RS Hi, You drew the line as if it is horizontal but I don't want to make that assumption. Here is my diagram. I called the line L. Reflect the point B in the line L to obtain a new point B'. Join A and B'. Let K be the point where this line meets L. Claim: The triangle ABK has the least perimeter of any triangle ABX where X is on L. Since KB = KB' the perimeter of ABK is AB + AB'. If X is any other point on L then XB = XB' and the perimeter of ABX is AB + AX + XB'. But AX + XB' > AB'. We are not sure what you are trying to do but if you are trying to construct the point K here is one procedure. Put a coordinate system on the plane. Let A = (a_{1}, a_{2}), B = (b_{1}, b_{2}) and L be the line y = mx + b.
Chris and Penny  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 