Radius of Circumscribed Circle

Date: 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/