Radius of Circumscribed CircleDate: 05/11/2001 at 12:57:31 From: Peter W. Murphy Subject: Derive R = abc/4K for the radius of circumscribed circle On your page on Triangle Formulas in the Dr. Math FAQ at http://mathforum.org/dr.math/faq/formulas/faq.triangle.html you give the formula R = abc/4K for the radius of the circle circumscribed about the scalene triangle abc, where K is the area of that triangle (obtainable, e.g., from Heron's formula). Where (online, or if necessary in the literature) can I find a derivation of R = abc/4K? Or can you provide one yourself? (I recognize that geometry in an essentially ASCII medium is tedious.) I am interested in a derivation because I am an assistant MathCounts coach and would like to have answers for bright kids. Thanks (and thanks for a great Web site!), Peter W. Murphy Livermore, CA Date: 05/11/2001 at 13:44:15 From: Doctor Schwa Subject: Re: Derive R = abc/4K for the radius of circumscribed circle Hi Peter, The derivation that I know is a bit above the level a MathCounts kid could appreciate: they don't generally know any trigonometry yet. I start with the area K = 1/2 ab sin C, and then use the law of sines, c/sin C = 2R to get sin C = c/2R, whereupon K = abc/4R. I don't know of any more elementary geometric proofs of this fact. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 12/03/2002 at 09:57:44 From: Gary B. Hicks Subject: Your derivation of the radius of a circumscribed circle In your solution for the radius of a circumscribed circle about a triangle, you magically stated that because of the Law of Sines, sin C = c/(2r). I could not derive that from the Law of Sines, but rather from the fact that when radii are drawn to the vertices, the base triangle with side C has an acute angle which is the complement of the vertex angle C. This means that sin C = cos (base angle of lower triangle) which is 1/2(c)/r, and thus the substitution. I would be interested in seeing your derivation from the Law of Sines. G. Hicks, Teacher, Dunbar High School Fort Worth, TX Date: 12/03/2002 at 11:32:29 From: Doctor Schwa Subject: Re: Your derivation of the radius of a circumscribed circle Since the law of sines says a / sin A = b / sin B = c / sin C = 2R it's pretty easy to derive! I know a lot of books leave off the "= 2R" part, and some even go so far as to put the sines on top instead of on the bottom. I think the best way to prove the law of sines is to prove that a / sin A = 2R and then by symmetry the rest follows. So I am pretty insistent on getting that 2R in there. Of course your proof that sinC = c/(2R) is equivalent to proving the law of sines (when you supplement it with the symmetry argument to show that it must also be true for B and A). By the way, usually we use R for the circumradius and r for the inradius to avoid confusion. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 12/04/2002 at 06:26:28 From: Gary B. Hicks Subject: Your derivation of the radius of a circumscribed circle I have never seen a statement of the Law of Sines that included "twice the radius of the circumscribed circle." However, it is true, and thank you for bringing it to my attention. My concern was how you arrived at it. If we started the proof of the Law of Sines by saying "It is intuitively obvious to the most casual observer that since any triangle has a circumscribed circle, twice the radius of the circle is equal to any side divided by the sin of the opposite angle, and hence c/sinC = a/sinA = b/sinB," do you not think that the reader would wonder how you arrived at the hypothesis? Maybe not on your plateau, but certainly on mine. :) Date: 12/04/2002 at 07:27:03 From: Doctor Schwa Subject: Re: Your derivation of the radius of a circumscribed circle My proof is just like yours: I draw in a radius of the circle, and show that (c/2) / sin C = R, and thus c / sin C = 2R, and then by symmetry, the law of sines is proved. I think this is a good argument, especially for students like mine where sin and cos are introduced as circular functions in the first place: why not use the circle? Of course the area argument, 1/2 a b sin C = 1/2 b c sin A, is not bad either. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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