INRADIUS 1

Let P be a point, 1,2,3,4 four lines passing through P and 0 a line intersecting the four lines at four distinct and real points (ie not passing through P and not parallel to some one of the four lines)

Denote:

r_ij := the inradius of the triangle bounded by the lines (0,i,j)

THEOREM

1/r_14 =

[(1/(r_12*r_24)) - (1/(r_13*r_34))] /

[((1/r_12) + (1/r_24)) - ((1/r_13) + (1/r_34))]

Simple application of the altitude formula found HERE.

Exercise for the reader:
Find the formula of the r_23

CONCURRENT EXTERNAL TANGENTS OF THREE CIRCLES

LEMMA

Let ABC be a triangle and D a point on BC.
Assume that D is between B and C.

Let r_a, r_b, r_c be the radii of the incircles
of the triangles ABC, BAD, CAD, resp.

And R_a, R_b, R_c the radii of the corresponding excircles, ie
Ra := the a-exradius of the triangle BAC
Rb := the a-exradius of the triangle BAD
Rc := the a-exradius of the triangle CAD.

This equality holds:

(r_b / R_b) * (r_c / R_c) = r_a / R_a

Proof 1 (trigonometrically):

In every triangle we have that

Inradius r = 4Rsin(A/2)sin(B/2)sin(C/2)

Exradius r_1 = 4Rsin(A/2)cos(B/2)cos(C/2)

By applying these formulae to the triangles ABC, ABD, ACD, we
get:

r_a / R_a = tan(B/2)tan(C/2)

r_b / R_b = tan(BDA/2)tan(B/2)

r_c / R_c = tan(CDA/2)tan(C/2)

==> (r_b / R_b) * (r_c / R_c) = r_a / R_a

since tan(BDA/2)tan(CDA/2) = 1 (since BDA+CDA = Pi)

Proof 2 (algebraically):

Denote

AB = c, BC = a, CA = b, AD = z, BD = x, DC = y

In every triangle ABC we have that:

Inradius r = area(ABC) / s

Exradius r_1 = area(ABC) / s-a

(where s = semiperimeter of ABC)

Now, by applying these formulae in the triangles
ABC, ABD, ACD, the formula to prove:
(r_b / R_b) * (r_c / R_c) = r_a / R_a

becomes:

(c+z-x)/(c+z+x) * (b+z-y)/(b+z+y) = (-a+b+c) / (a+b+c) ==>

abc + abz + acz + az^2 + axy =

= b^2x + bzx + bcy + byz + bcx + czx + c^2y + cyz (#)

By the Stewart Theorem we get:

b^2x +c^2c = z^2a + axy

The (#) becomes:

abc + abz + acz = bzx + bcy + byz + bcx + czx + cyz

and by replacing x with a-y (since x+y = a), we finally get:

0 = 0

THEOREM

If h_a is the common altitude (from A) of the triangles
ABC, ABD, ACD, then

1. h_a = 2r_b*r_c / (-r_a + r_b + r_c)

2. h_a = 2R_b*R_c / (R_a - R_b - R_c)

In every triangle:

2/h_a = 1/r - 1/r_1 (where r,r_1 are the inradius, a-exradius,resp.)

By applying the formula to the triangles ABC, ABD, ACD,
we get:

2/h_a = 1/r_a - 1/R_a = 1/r_b - 1/R_b = 1/r_c - 1/R_c

From these equalities we get:

R_a = h_a * r_a / (h_a - 2r_a)

R_b = h_a * r_b / (h_a - 2r_b)

R_c = h_a * r_c / (h_a - 2r_c)

By replacing the Ra,Rb,Rc in the

(r_b / R_b) * (r_c / R_c) = r_a / R_a

we get:

h_a = 2r_b*r_c / (-r_a + r_b + r_c)

Similarly, from the equalities above, we get:

r_a = h_a * R_a / (h_a + 2R_a)

r_b = h_a * R_b / (h_a + 2R_b)

r_c = h_a * R_c / (h_a + 2R_c)

and by replacing the r_a,r_b,r_c in the

(r_b / R_b) * (r_c / R_c) = r_a / R_a

we get

h_a = 2R_b*R_c / (R_a - R_b - R_c)

TRIANGLE CONSTRUCTION a, B - C, h_b + h_c

To construct triangle ABC if are given a, B - C, h_b + h_c, where h_b, h_c are the altitudes BB', CC'.

Solution 1.

Analysis:

Let ABC be the triangle with BC = a, B-C, BB' + CC' = h_b + h_c given.

Let D be the point on CA such that AD = AB with the A between D,C
(DC = DA + AC = AB + AC).

The triangle ABD is isosceles with ang(ADB) = ang(ABD) = A/2.

Let E be the orthogonal projection of C on DB.

In the triangle EBC we have:

ang(CEB) = 90 d., BC = a, ang(EBC) = ang(BCD + BDC) = C + (A/2) =
90 - ((B-C)/2) d.

Therefore EC is known, since the triangle EBC can be constructed.

From the similar triangles ECD and B'BD we have:

EC / DC = BB' / BD or

EC / (AB + AC) = BB' / DB (1)

From the similar right triangles ABB', ACC' we have:

BB' / AB = CC' / AC ==>

BB' / AB = CC' / AC = (BB' + CC') / (AB+AC) ==>

BB' = AB(BB' + CC') / (AB + AC) (2)

From (1) and (2) we get:

EC / (BB' + CC') = AB / DB : fixed,
since EC and BB' + CC' are known.

Therefore the triangle ADB "remains similar to itself", that is, it has known angles. Therefore angle A is known, since ang(BDA) = ang(ABD) = A/2

Construction: We leave it to the reader.

Solution 2.

Analysis:

Let h_a = AD, AE be the altitude, angle bisector from A, resp.

The parallel from E to AB intersects AC at Z.

The triangle ZAE is isosceles with EZ = AZ
(since angles ZEA = BAE = EAZ = A/2)

EZ / AB = CZ / CA = (CA - AZ) / CA = (CA - EZ) / CA

==> 1/EZ = 1/b + 1/c (1)

2*area(ABC) = ah_a = bh_b = ch_c ==>

h_2 + h_3 = ah_a (1/b + 1/c) (2)

From (1) and (2) we get that

h_a / EZ = AD / EZ = (h_b + h_c) / a : fixed (3)

The triangle DAE has known angles:

ang(ADE) = 90 d., ang(DAE) = (B-C)/2, (DEA) = 90 - ((B-C)/2)

Therefore AD / AE = h_a / AE is fixed. (4)

From (3) and (4) by division we get that:

AE / EZ is fixed.

Now, in the isosceles triangle ZAE we have that AE / EZ is fixed, therefore "it remains similar to itself", that is its angles are known. So A = 2*ang(EAZ) is known.

Construction: We leave it to the reader.

Problem for the reader:
To construct ABC if are given a, B-C, h_c - h_b.