Date: 04/16/2002 at 06:39:22 From: Louise Hjarnaa Subject: Classical geometry Let ABC be a triangle with sides a, b, c. Let h be the perpendicular from A to a, and m the median from A to the midpoint of a. How can I construct the triangle - in a classical way using only ruler and compass - when I know A, h, m ?
Date: 04/18/2002 at 17:05:23 From: Doctor Floor Subject: Re: Classical geometry Hi, Louise, Thanks for your question. I was pointed to the following simple solution for this problem by my friend Antreas P. Hatzipolakis. Before giving the construction, we consider a triangle ABC, its midpoints Ma, Mb, and Mc, and the foot Ha of the altitude from A. A /| \ / | \ / | \ Mc--M-------Mb / | \ / | \ / | \ B----- Ha--Ma-----------C First of all we note that MbMc intersects AHa at its midpoint, and that MbMc and AHa are perpendicular. So MbMc are perpendicular bisectors. Then we note that MaMb is parallel to AB, so that angle MaMbA is equal to 180-<A. The locus of points X such that MaXA is equal to 180-<A is a circular arc. The circle containing this arc passes through A, Ma, and the top vertex of the isosceles triangle on base AMa with base angles <A/2 (which can be constructed). Thus our construction is as follows: * From the given lengths h and m, and the fact that angle AHaMa is right, construct AHaMa. * Construct the perpendicular bisector L of AHa. * Construct the isosceles triangle AMaD, with D as top vertex, and base angles <A/2, which are constructable since <A was given. There is a choice here. For instance, take D and Ha opposite to each other w.r.t. AMa. * Construct the circumcircle of AMaD, and only use the arc AMa containing D. We call this arc H. * Intersect H and L; this gives Mb (or Mc, which is not important). * Reflect A through Mb to get C. * Reflect C through Ma to get B. And we have ABC. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Date: 04/22/2002 at 10:52:26 From: Louise Hjarnaa Subject: Classical geometry Dear Doctor Floor and Antreas P. Hatzipolakis, It is a very nice solution - I am impressed - thank you. Best regards Louise
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum