Frequently Asked Questions about (Plane) Triangles.

Henry G. Baker

hbaker@netcom.com

October, 1996.

Copyright (c) 1996 by Henry G. Baker.  All rights reserved.

This FAQ file is located at Web URL:
ftp://ftp.netcom.com/pub/hb/hbaker/FAQ-triangle.txt

Q.  How to find the diameter of the 'circumcircle': the circle which
circumscribes a triangle?

I.e., Given a triangle with vertices A,B,C (with respective angles
A,B,C) and opposite sides a,b,c, respectively, find the diameter of
the circle which circumscribes the points A,B,C.

We make use of the following theorem:

THEOREM.  In the circle circumscribing points A,B,C, the central angle
of the arc BC (with chord a) is A/2, _independent_ of the actual
values of the angles B,C.

We then move point A while holding the position and the length of side
a constant, and holding angle A constant.  The locus of these points
is an arc on the circle which circumscribes A,B,C.

Let b', B', c', C' denote the new sides and new angles generated
during the motion of A along the circumscribed circle.  Note that we
do not need a' or A', since neither a nor A has changed.

At one extreme on the locus of these points, the new angle B' is zero,
while at the other extreme on the locus of these points, the angle B'
is 180 degrees.  Since the motion is continuous, there is a point on
this locus at which the angle B' is 90 degrees.  But in this case,
side b' becomes a diameter D of the circumscribed circle.

Now consider sin A = (opposite)/(hypoteneuse) = a/b'.  Rearranging, we
get b' = a/sin(A).  But b' is the diameter D of the circumcircle;
therefore D = b' = a/sin(A).

By symmetry, we can use exactly the same argument for the other sides
and angles.

We have now proved "the law of sines":

D = a/sin(A) = b/sin(B) = c/sin(C)

----------

Q.  How to find the area of a triangle in terms of the lengths of its
sides?

I.e., given a triangle with vertices A,B,C (with respective angles
A,B,C), and opposite sides a,b,c, respectively, find an expression for
the area of the triangle in terms of a,b,c.

Let V be the vector CB and W be the vector CA.  Then |V|=a and |W|=b.
Consider the cross product VxW of V,W.  Then

Area = 1/2 |VxW| = 1/2 |V| |W| sin(C) = 1/2 a b sin(C).

But by the law of sines, the diameter of the circumcircle D=c/sin(C),
sin(C)=c/D, so

Area = 1/2 a b sin(C) = 1/2 a b c/D = abc/(2D) = abc/(4R),
where R = radius of the circumcircle.

We thus have expressed the triangle area in terms of the sides and the
radius R of the circumcircle.

Consider now the circle _inscribed_ within the triangle A,B,C: the
'incircle'.  The center of the incircle lies at the intersection of
the angle bisectors for the three vertices of the triangle.

We can now break up the original triangle in to 3 little triangles
having the original sides as bases, and whose third vertices are all
identically the center of the incircle.  We note that the _altitudes_
of all three triangles are identical, and equal to the radius r of the
incircle.

The area of the original triangle can now be obtained by summing the
areas of the three little triangles:

Area = 1/2 a r + 1/2 b r + 1/2 c r = 1/2 (a+b+c) r = s r, where s =
(a+b+c)/2 is the 'semiperimeter' of the original triangle, and r is
the radius of the incircle.

We then break each of the smaller triangles at the altitude into 2
right triangles, and we notice that there are two congruent triangles
on either side of the angle bisectors of the original triangle.  We
now consider the lengths of the bases of these 6 smaller triangles (3
congruent pairs).  We denote the bases of the congruent pair adjacent
to the vertex A by x, the bases of the congruent pair adjacent to the
vertex B by y, and the bases of the congruent pair adjacent to the
vertex C by z.  We note that

a = y+z
b = x+z
c = x+y

This is three linear equations in three unknowns, and if we solve
them, we get

2x = -a+b+c, or x = s-a
2y =  a-b+c, or y = s-b
2z =  a+b-c, or z = s-c

Incidentally, we have now computed the following tangents:

tan(A/2) = r/(s-a)
tan(B/2) = r/(s-b)
tan(C/2) = r/(s-c)

If we now sum up the areas of these 6 small triangles (3 congruent
pairs), we get

2 1/2 x r + 2 1/2 y r + 2 1/2 z r = xr+yr+zr
                                  = (x+y+z)r
                                  = (s-a+s-b+s-c)r
                                  = (3s-2s)r
                                  = sr

By equating these two expressions for the area of a triangle, we can
get a relationship between the radius R of the circumcircle and the
radius r of the incircle.

Area = sr = abc/(4R), so

4Rr = abc/s, or

2Rr = abc/(a+b+c)

We would now like to express tan(A/2) directly in terms of a,b,c.

Using the 'cosine' formula, we can compute cos(A):

a^2 = b^2 + c^2 - 2bc cos(A), or

cos(A) = (b^2 + c^2 - a^2)/(2bc)

Using the fact that a = 2s-(b+c), we get

cos(A) = 1 - 2(s-b)(s-c)/(bc)

We now recognize this as the cosine double-angle formula:

cos(A) = 1 - 2(s-b)(s-c)/(bc) = 1 - 2 sin^2(A/2), so

sin^2(A/2) = (s-b)(s-c)/(bc), and therefore

sin(A/2) = sqrt((s-b)(s-c)/(bc))

Now we can compute tan(A/2) as:

tan(A/2) = sin(A/2)/sqrt(1-sin^2(A/2))
         = sqrt((s-b)(s-c)/(bc)) / sqrt(1-(s-b)(s-c)/(bc))
         = sqrt((s-b)(s-c)) / sqrt(bc-(s-b)(s-c))
         = sqrt((s-b)(s-c)) / sqrt(s(s-a))
         = sqrt((s-b)(s-c)/(s(s-a)))

Using the fact (proved above) that tan(A/2)=r/(s-a), we now get the
radius r of the incircle in terms of the sides a,b,c:

r = (s-a) sqrt((s-b)(s-c)/(s(s-a)))
  = sqrt((s-a)(s-b)(s-c)/s)

Since the area of the triangle is sr, we have

Area = sr = s sqrt((s-a)(s-b)(s-c)/s) = sqrt(s(s-a)(s-b)(s-c)), which
is "Heron's formula" for the area of a triangle in terms of the sides
a,b,c.

The radius R of the circumcircle in terms of the sides a,b,c is now

R = abc/(4rs) = 1/4 abc/sqrt(s(s-a)(s-b)(s-c))

-----