Constructing a Triangle Given the Medians

Date: 01/01/2001 at 14:32:24
From: Ali
Subject: Geometry

Dr. Math:

How can I construct a triangle ABC given AM, BN, and CP, the 
respective medians from the vertices A, B, and C? 

For construction, assume the lengths, angles or whatever is given.  

Can you please tell me what the construction steps are? If there are 
limitations on the givens, what are those limitations?

Date: 01/03/2001 at 12:43:42
From: Doctor Rob
Subject: Re: Geometry

Thanks for writing to Ask Dr. Math, Ali.

Re-label so that AM is the longest of the three medians.

Draw line segment AM, and trisect it with points Q and R so that 
AR = RQ = QM. Draw a line segment SV with length the same as that of 
BN, and trisect it with points T and U so that ST = TU = UV. Draw a 
line segment WZ with the same length as that of CP, and trisect it 
with points X and Y so that WX = XY = YZ.

Using point R as center, draw a circle with radius ST. Using point Q 
as center, draw a circle with radius YZ. Then point P will lie at one 
of the two intersections of these two circles. (Either one will do.) 
Draw line PQ and locate point C beyond Q on it. Connect AC. Find the 
midpoint of AC, N. Draw line NQ and locate point B beyond Q on it. 
Connect AB and BC.

All that is required is that the two circles intersect in two points. 
That requires that the distances between their centers be less than 
the sum of their radii. In symbols, that means:

     QR < ST + YZ
     AM/3 < BN/3 + CP/3
     AM < BN + CP

This is just the triangle inequality. That means that the only 
restriction on the lengths of the medians is that they themselves can 
be the lengths of the sides of a triangle.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/