Inscribed circle

[onta...@hotmail.com]

Find the equation of the inscribed circle of the triangle formed by the lines
4x-3y+9 = 0
12x+5y-43 = 0 and
7x+24y+19 = 0

[shy author]

Let the lines be labeled:

(a) 4x-3y+9 = 0
(b) 12x+5y-43 = 0
(c) 7x+24y+19 = 0

Using Gaussian solution of MX=B for matrices/vectors, we find the
vertices of the triangle to be:

A = [49/11, -23/11] (intersection of lines b,c)
B = [-7/3, -1/9] (intersection of lines a,c)
C = [3/2, 5] (intersection of lines a,b)

And the opposite side lengths are

a = 115/18
b = 169/22
c = 700/99

which satisfies the triangle inequalities on all 3 sides.

The area of this triangle, using Heron's formula is

area = 2093/99

showing that we do have a Heron triangle.

The radius of the incircle = 2*area/(a+b+c) = 2

The centre of the incircle is at

[ (a*A[1]+b*B[1]+c*C[1])/(a+b+c), (a*A[2]+b*B[2]+c*C[2])/(a+b+c) ]

which is [1,1]

For the incircle with centre at [1,1] and radius 2, we have the equation

(1) (x - 1)^2 + (y - 1)^2 = 2^2 = 4

which is what we seek.

Things get much more interesting if we have a triangle in 3d space.

[onta...@hotmail.com]

Not quite.

[shy author]

On Sat, 12 Jul 2008 15:21:41 +0000, onta...@hotmail.com wrote:

> Not quite.

did I make a simple math mistake again?

[onta...@hotmail.com]

Not a big deal !

x^2+y^2-2x-2y-2 = 0

[Dan Cass]

To which the OP responded "not quite".
But the required circle is indeed centered at (1,1)
and has radius 2. Perhaps the OP is objecting
to something about the method?
Anyway here's another approach.
First rewrite the equations so the y coeffs are positive.
L1 -4x + 3y - 9 = 0
L2 12x + 5y - 43 = 0
L3 7x + 24y + 19 = 0.
A graph shows the center of the inscribed circle is above
line L3 and below lines L1 and L2.
So the center is at the place where
-dist(P,L1) = -dist(P,L2) = dist(P,L3).
Here for any point X and line L dist(X,L) is the
signed distance from X to line L which is positive for
points above line L. In general for a line with equation
ax+by+c=0 and b>0 this signed distance for P(x,y) is
.. [ax+by+c)/sqrt(a^2+b^2).
Applying this to the present problem, the following are to be equal:
-(-4x+3y-9)/5 = -(12x+5y-43)/13 = +(7x+24y+19)/25.
This has unique solution (x,y) = (1,1) at which
ass three signed distances are 2, so the required
circle is as in previous response centered at (1,1) and
has radius 2. That the equation is
.. (x-1)^2 + (y-1)^2 = 4
may be seen after 3 pages of careful computation.

[Dan Cass]

> ... [ax+by+c)/sqrt(a^2+b^2).

> Applying this to the present problem, the following
> are to be equal:
> -(-4x+3y-9)/5 = -(12x+5y-43)/13 = +(7x+24y+19)/25.
> This has unique solution (x,y) = (1,1) at which
> ass three signed distances are 2, so the required
> circle is as in previous response centered at (1,1)
> and
> has radius 2. That the equation is

> ... (x-1)^2 + (y-1)^2 = 4

> may be seen after 3 pages of careful computation.

I should have said, "at which all three", rather than
"at which ass three"... my bad.

[onta...@hotmail.com]

Do you have a general equation then for an inscribed sphere, given the equations of the four faces of an irregular tetrahedron ? That would be interesting.

[BSK]

A1*x + B1*y + C1 = 0, A1^2 + B1^2 = 1
A2*x + B2*y + C2 = 0, A2^2 + B2^2 = 1
A3*x + B3*y + C3 = 0, A3^2 + B3^2 = 1

The following method works for any dimension

First, find X and Y from
(a,a)X + (a,b)Y +(a,c) = 0
(b,a)X + (b,b)Y +(b,c) = 0
where
(a,a) = A1*A1 + A2*A2 + A3*A3
(a,b) = A1*B1 + A2*B2 + A3*B3
..

Then the center (x, y), and the radius (R) find from
A1*x + B1*y + C1 = R*sign(A1*X + B1*Y + C1)
A2*x + B2*y + C2 = R*sign(A2*X + B2*Y + C2)
A3*x + B3*y + C3 = R*sign(A3*X + B3*Y + C3)

[Avni Pllana]

> Do you have a general equation then for an inscribed
> sphere, given the equations of the four faces of an
> irregular tetrahedron ? That would be interesting.

Hi Bill,

We talked about a triangle in 3D, and in this case we proceed exactly as in 2D. Barycentric coordinates work similarly on tetrahedra. Let A, B, C, D, be the vertices of the tetrahedron, and Sa, Sb, Sc, Sd, be the areas of their respective faces. The incenter It of tetrahedron is determined as follows:

It = (Sa*A + Sb*B + Sc*C + Sd*D)/(Sa+Sb+Sc+Sd) .

The radius R of the inscribed sphere is

R = 3*V/(Sa+Sb+Sc+Sd) ,

where V is the volume of the tetrahedron.

Best regards,
Avni

[BSK]

> Your idea is quite good in the particular case of
> the incircle, as shown by BSK, but shy author used
> the very powerful technique of barycentric
> coordinates to solve the problem.

Not in the particular case. The idea works in general case and requires only one matrix by matrix multiplication and solution of two linear systems.

Whereas with the very powerful technique of barycentric coordinates one have to find vertices, areas, volumes ...

[Avni Pllana]

The incenter is just one among the large number of interesting triangle centers. For example try to determine the Nagel Point of the triangle with given vertices A, B, C.

Best regards,
Avni
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> The incenter is just one among the large number of
> interesting triangle centers. For example try to
> determine the Nagel Point of the triangle with given
> vertices A, B, C.
>
> Best regards,
> Avni

My method finds this just one among the large number.