Relations between various elements of a triangle
S = rp
Triangle ABC is a union of three triangles ABI, BCI, CAI, with bases AB = c, BC = a, and AC = b, respectively. The altitudes to those bases all have the length of r.
r2 = p-1(p - a)(p - b)(p - c)
This follows from S2 = p(p - a)(p - b)(p - c) and S = rp.
1/r = 1/ha + 1/hb + 1/hc
2S = aha = bhb = chc. Therefore, a = 2S/ha, etc. On the other hand, S = rp, so that p = S/r, or (a + b + c) = 2S/r. From here, 2S/ha + 2S/hb + 2S/hc = 2S/r.
sin2(A/2) = (p - b)(p - c) / bc, etc.
First of all, sin(A) = 2·sin(A/2)cos(A/2) = 2·sin2(A/2)/tan(A/2). Therefore,
| (1) |
sin2(A/2) = sin(A)·tan(A/2) /2.
|
We know that
and
Combining (1)-(3) gives
| |
sin2(A/2) = 2S/bc · r/(p-a) · 1/2.
|
Taking into account that S2 = p(p - a)(p - b)(p - c) and r2 = p-1(p - a)(p - b)(p - c), the latter leads to
| |
sin2(A/2) = (p - b)(p - c) / bc.
|
AI2 = (p - a)bc/p
Square the obvious
Substitute there sin2(A/2) = (p - b)(p - c) / bc and r2 = p-1(p - a)(p - b)(p - c):
| |
| AI2 | = p-1(p - a)(p - b)(p - c)bc/(p - b)(p - c) |
| | = (p - a)bc/p. |
|
bc·tan(B/2)·tan(C/2)
Squaring AI = r/sin(A/2) and substituting sin2(A/2) = (p - b)(p - c) / bc, we obtain
| |
AI2 = r2·bc/(p - b)(p - c).
|
By the incenter construction, tan(B/2) = r/(p - b) and also tan(C/2) = r/(p - c). Substituting these into the above gives the required
| |
AI2 = bc·tan(B/2)·tan(C/2).
|
1/r = 1/ra + 1/rb + 1/rc
As we know,
| |
S = ra(p - a) = rb(p - b) = rc(p - c).
|
Therefore
| |
| 1/ra+ 1/rb + 1/rc | = (p - a)/S + (p - b)/S + (p - c)/S |
| | = (3p - a - b - c)/S |
| | = (3p - 2p)/S |
| | = p/S |
| | = 1/r, |
|
since S = rp.
ra + rb + rc = r + 4R
As we know,
and also
| |
S = ra(p - a) = rb(p - b) = rc(p - c).
|
From these we have
| (4) |
ra + rb + rc - r = S(1/(p - a) + 1/(p - b) + 1/(p - c) - 1/p).
|
Simple algebra yields
| |
1/(p - a) + 1/(p - b) = c / (p - a)(p - b) and
1/(p - c) - 1/p = c / p(p - c).
|
And a little more effort makes a great payoff:
| |
c / (p - a)(p - b) + c / p(p - c) = abc / p(p - a)(p - b)(p - c) = abc / S2,
|
by Heron's formula. To sum up, from (4)
| (5) |
ra + rb + rc - r = S·abc/S2 = abc / S.
|
However, abc = 4RS, so that (5) implies exactly what's needed:
| |
ra + rb + rc - r = abc / S = 4R.
|
rarbrc = pS
Since
| |
S = ra(p - a) = rb(p - b) = rc(p - c),
|
we immediatly obtain
| |
| rarbrc | = S3 / (p - a)(p - b)(p - c) |
| | = S3 / [S2 / p], |
|
by Heron's formula. But
r rarbrc = S2
This is an immediate consequence of rarbrc = pS and rp = S.
ma2 = (b2 + c2)/2 - a2/4
Let's use Stewart's theorem
| |
AB2·DC + AC2·BD - AD2·BC = BC·DC·BD
|
with D being the midpoint M of BC. Then AB = c, DC = a/2, AC = b, BD = a/2, AD = ma, BC = a. We have,
| |
c2·a/2 + b2·a/2 - ma2·a = a·a/2·a/2.
|
(The above identity could be as easily obtained with the help of the theorem of Cosines.)
abc = 4RS
Let AD be a diameter of the circumcircle of  ABC and AH its altitude. Right triangles AHC and ABD are similar, for  ADB = ACH. Therefore,
In other words,
And finally
bc = 2Rha
This follows from the previous derivation or by substituting S = aha/2 into the final formula.
p = 4Rcos(A/2)·cos(B/2)·cos(C/2)
By the Law of Sines
| |
a = 2R·sinA, b = 2R·sinB, c = 2R·sinC,
|
so that
| |
| p | = R·(sinA + sinB + sinC) |
| | = R·(sinA + sinB + sin(180° - A - B) |
| | = R·(sinA + sinB + sin(A + B) |
| | = R·(sinA + sinB + sinA·cosB + cosA·sinB) |
| | = R·(sinA·(1 + cosB) + sinB·(1 + cosA)) |
| | = R·(2sin(A/2)cos(A/2)·2cos2(B/2) + 2sin(B/2)cos(B/2)·2cos2(A/2)) |
| | = 4R·cos(A/2)cos(B/2)(sin(A/2)cos(B/2) + sin(B/2)cos(A/2)) |
| | = 4R·cos(A/2)cos(B/2)sin((A + B)/2) |
| | = 4R·cos(A/2)cos(B/2)sin(90° - C/2) |
| | = 4R·cos(A/2)cos(B/2)cos(C/2). |
|
S = 2R2sin(A)·sin(B)·sin(C)
By the Law of Sines
| |
a = 2R·sinA, b = 2R·sinB,
|
For the area of the triangle we have
| |
| 2S | = ab·sinC |
| | = 2RsinA·2RsinB·sinC |
| | = 4R2·sinA·sinB·sinC. |
|
r = 4Rsin(A/2)·sin(B/2)·sin(C/2)
This follows directly from
- S = rp,
- p = 4Rcos(A/2)·cos(B/2)·cos(C/2), and
- S = 2R2sin(A)·sin(B)·sin(C).
cot(A/2) + cot(B/2) + cot(C/2) = cot(A/2)·cot(B/2)·cot(C/2)
This is equivalent to showing that, for A + B + C = 180°,
| | cos(A/2)sin(B/2)sin(C/2) + sin(A/2)cos(B/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2)
= cos(A/2)cos(B/2)cos(C/2).
|
Let's transform the left-hand side:
| |
cos(A/2)sin(B/2)sin(C/2) + sin(A/2)cos(B/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2)
= sin((A+B)/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2).
|
But since (A + B)/2 C = 90° - C/2, this equals
| |
cos(C/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2) = cos(C/2)[sin(C/2) + sin(A/2)sin(B/2)].
|
Reversing the steps:
| |
| sin(C/2) + sin(A/2)sin(B/2) | = cos((A+B)/2) + sin(A/2)sin(B/2) |
| | = cos(A/2)cos(B/2) - sin(A/2)sin(B/2) + sin(A/2)sin(B/2) |
| | = cos(A/2)cos(B/2). |
|
Combining everything together we get the desired identity.
rR = abc / 4p
r2 = p-1(p - a)(p - b)(p - c) is equivalent to
where D = p(p - a)(p - b)(p - c). Also,
Multiplying the two gives
Copyright © 1996-2008 Alexander Bogomolny
| 
