Constructing a Triangle Given the MediansDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/