Triangle Construction Given an Angle, the Inradius, and the SemiperimeterDate: 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/ |
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