Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Classical Geometry

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
Associated Topics:
College Constructions
College Triangles and Other Polygons
High School Constructions
High School Triangles and Other Polygons

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/