A trigonometrical problem:?

The marked angles are equal by the Inscribed Angle Theorem.

(https://en.wikipedia.org/wiki/Inscribed_angle)

Note that by similar triangles,

h . . .  c/2

--- = -------.................................. (1)

b. . . . . r

Thus, the area of the triangle is

S  =  ah/2  =  a*b*c/4 r......................(2)

Therefore, the circum-radius is

r = a b c/ 4 S...................................(3)   ................... Proved

My way:

the sine rule gives

a/sinA=b/sinB=c/sinC=2r

=>

(a+b+c)/2=r(sinA+sinB+sinC)

=>

s=r(2S/bc+2S/ac+2S/ab),

where

[s=(a+b+c)/2 &

 S=absinC/2=bcsinA/2=casinB/2;

 sinA=2S/bc,sinB=2S/ac,sinC=2S/ab]

=>

s=2rS(1/bc+1/ac+1/ab)

=>

s=2rS(a+b+c)/abc

=>

s=4rSs/abc

=>

r=abc/(4S).

this is a type of problem which confuses a student.

suppose

area of the triangle = S sq.cm

radius of the circumcircle = R cm

how can length and area be equal?

though numerically same, units are different.

all that has been achieved is manipulation of

the numbers.

We want to prove r = abc/(4S).

S is the area of △ABC (BC = a, AC = b, AB = c), so

S = (1/2)ab*sin(∠ACB).

Let the center of the circle be O.

∠ACB = ∠AOB/2, so

S = (1/2)ab*sin(∠AOB/2)

sin(∠AOB/2) can not be negative, so

sin(∠AOB/2) = √[1 - cos^2(∠AOB/2)]

We know cos(2θ) = 2cos^2(θ) - 1 so cos^2(θ) = (cos(2θ) + 1)/2.

Therefore

sin(∠AOB/2) = √[1 - (cos(∠AOB) + 1)/2]

 = √[1/2 - cos(∠AOB)/2]

So

S = (1/2)ab√[1/2 - cos(∠AOB)/2]

Think △AOB. AO = BO = r and AB = c, so by the cosine law,

c^2 = r^2 + r^2 - 2r^2*cos(∠AOB)

cos(∠AOB) = (2r^2 - c^2)/(2r^2)

 = 1 - c^2/(2r^2)

So

S = (1/2)ab√[1/2 - (1 - c^2/(2r^2))/2]

 = (1/2)ab√[c^2/(4r^2)]

 = (1/2)ab(c/(2r))

 = abc/(4r)

Therefore

abc/(4S) = abc/(abc/r) = r

This is how I did it: consider it the other way around  ;)

Via scaling, consider the unit circle. WLOG, Let A = (1,0). B & C the other two points.

Now your problem becomes a simple two parameter system:  B = (d, √(1-d²)) and C = (e, √(1-e²)).  Using pythagora, areas (all basic geometry) your result follows after a little algebra...

The center, O, of the circumcircle is equidistant from all 3 vertices.

Given:

BC = a

AC = b

AB = c

OA = OB = OC = r

Draw a diameter BD that goes through O. 

OD = r

BD = 2r

Also drop a perpendicular BE from B down to AC. 

BE = h

∠BEC is a right angle by the definition of perpendicular.

∠BAD is a right angle by Thales' Theorem (BD is a diameter)

∠ADB and ∠BCA subtend the same arc AB, so they are congruent.

By the AA Similarity Postulate, △BAD ~ △BEC.

Corresponding parts of those triangles are proportional so:

BD/BA = BC/BE

Substituting we have:

2r/c = a/h

The area S, of △ABC is:

S = bh/2

Solving for h:

h = 2S/b

Substituting for h:

2r/c = a / (2S/b)

2r/c = ab / (2S)

Solving for r:

r = ab / (2S) * c/2

r = abc / (2*2S)

r = abc / (4S)

Q.E.D.