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Triangle Construction Given an Angle, the Inradius, and the Semiperimeter


Date: 03/26/2002 at 10:54:08
From: Tiffany
Subject: Triangle construction

Given an angle, alpha, the inradius (r), and the semi-perimeter (s) 
construct the triangle.


Date: 03/28/2002 at 07:50:17
From: Doctor Floor
Subject: Re: Triangle construction

Hi,

Thanks for your question.

Start with a circle with radius r; let's call the center M. From M 
draw two radii that make an angle of 180 degrees minus alpha. Let N1 
and N2 be the points where these radii meet the circle. The tangents 
to the circle at N1 and N2 meet each other in a point, say A. 

Now, in the Dr. Math archives at:

  Geometric Proof of Heron's Formula
  http://mathforum.org/library/drmath/view/54686.html   

in the first reply item 1., it is explained that, translated to our 
situation, AN1 = s-a (where a is the length of side BC opposite A).

Since we are given s as well, we easily have the length of BC = a.

Now note that the circumcenter O together with vertices BC forms an 
isosceles triangle with top angle 2*alpha (because the central angle 
of the circumcircle on BC is twice inscribed angle on BC). So we can 
construct this isosceles triangle, which gives us the radius R of the 
circumcircle.

Now we have the radii R and r and sidelength a. The construction from 
here is explained in the Dr. Math archives at:

  Triangle Construction
  http://mathforum.org/library/drmath/view/51843.html   

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 03/28/2002 at 11:32:28
From: Tiffany
Subject: Triangle construction

Thanks, 

Everything makes sense but I am not sure how to find the circumcenter 
O when I am not sure of the placement of BC. 

Tiffany


Date: 03/28/2002 at 16:16:02
From: Doctor Floor
Subject: Re: Triangle construction

Hi again, Tiffany,

Thanks for your reaction.

There is no need to find the circumcenter O. I just took a segment of 
the length of BC, and constructed O as if that were the real BC. That 
gave the actual length OB = R, which was all we needed.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Constructions
High School Geometry
High School Triangles and Other Polygons

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