Σάββατο, 25 Δεκεμβρίου 2010

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

Τετάρτη, 22 Δεκεμβρίου 2010

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)

Παρασκευή, 17 Δεκεμβρίου 2010

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.

Τρίτη, 20 Ιουλίου 2010

THREE CONCURRENT CIRCLES


5. Let 123 be a triangle, 4 a point inside 123 and (1') the circle touching the circles (134), (124) externally and the circle (234) internally at 5, (2') the circle touching the circles (214),(234) externally and the circle (314) internally at 6 and (3') the circle touching the circles (324),(314) externally and the circle (124) internally at 7.

The circles (167),(275),(356) concur at a point 8.

Variation:


Let (0) be the circle touching internally the circles (234), (314), (124) at 5,6,7 resp.


The circles (167),(275),(356) concur at a point 8.

Note:


If 4 is not inside triangle 123, but in the negative side of 23 (ie the side not containing 1), then (1') is the circle touching (314),(124) internally and (234) externally at 5.

Σάββατο, 3 Ιουλίου 2010

THREE CONCURRENT CIRCLES

Let 123 be a trianle, 4 a point, and 5,6,7 three points on the circles (423), (431),(412) resp. other than point 4.

Theorem:

The circles (167), (275) and (356) concur at a point (say) 8.


Special Cases:

1. Let the points 4,5,6,7 be concyclic (or collinear). See previous post.

2. Let the points 5,6,7 be the second intersections [= other than the point 4] of the lines 14, 24, 34 with the circles (423),(431),(412) resp.


3. Let the points 5,6,7 be the second intersections [= other than the point 4] of the circles (1,14),(2,24),(3,34) with the circles (423),(431), (412) resp.


4. Let 1',2',3' be the centers of the circles (423),(431),(412), resp. and the points 5,6,7, the second intersections [= other than the point 4] of the circles (42'3'),(43'1'),(41'2') with the circles (423),(431),(412) resp.


Continued 5

Παρασκευή, 2 Ιουλίου 2010

A SET ADDITION AND CONCURRENT CIRCLES

Let W = {a1,a2,a3,a4,....,an} be a finite set and S = {A1,A2,...,Ak} a set of subsets of W. We define the addition operation + in S:

X + Y = (X - (X ∩ Y)) U (W - (X U Y))

Example:

W = {1,2,3,4,5}, S = {{1,2,3},{1,3,4},{2,3,5}}

{1,2,3} + {1,3,4} = {2} U {5} = {2,5}

Examples from Geometry of closed sets S under the addition +
(ie if X, Y belong in S, then X + Y belongs in S as well).

FOUR CONCURRENT CIRCLES.

Let 123 be a triangle, 4 a point, 1', 2' and 3' the circles (234), (341) and (412), resp. and 4' an arbitrary circle passing through 4.

Denote:

5 := 4' ∩ 1' - {4}

6 := 4' ∩ 2' - {4}

7 := 4' ∩ 3' - {4}

(ie the other than point 4 intersections of the circle 4' with the circles 1',2',3', resp.)

5' := the circle (167), 6' := the circle (275), 7' := the circle (356)

Theorem:

The circles 5',6',7' concur at a point 8 on the circle
(123) := 8'


The circle 4' may be of infinite radius (ie be a line):


Consider the sets:

W = {1,2,3,4,5,6,7,8}, S = {1',2',3',4',6',7',8'}

where 1', 2',...,8' are sets of four concyclic points:
(not be confused with the circles 1', 2',... above:
1' above is the circle passing through the four points 2,3,4,5, while 1' is now the set containing the four points 2,3,4,5)

1' = {2,3,4,5}
2' = {1,3,4,6}
3' = {1,2,4,7}
4' = {4,5,6,7}
5' = {1,6,7,8}
6' = {2,5,7,8}
7' = {3,5,6,8}
8' = {1,2,3,8}

We have:

1' + 1' = {1,6,7,8} = 5'
1' + 2' = {2,5,7,8} = 6'
1' + 3' = {3,5,6,8} = 7'
1' + 4' = {1,2,3,8} = 8'
1' + 5' = {2,3,4,5} = 1'
1' + 6' = {1,3,4,6} = 2'
1' + 7' = {1,2,4,7} = 3'
1' + 8' = {4,5,6,7} = 4'

etc

Addition Table:

+ | 1' 2' 3' 4' 5' 6' 7' 8'
------------------------------------
1' | 5' 6' 7' 8' 1' 2' 3' 4'

2' | 5' 6' 7' 8' 1' 2' 3' 4'

3' | 5' 6' 7' 8' 1' 2' 3' 4'

4' | 5' 6' 7' 8' 1' 2' 3' 4'

5' | 5' 6' 7' 8' 1' 2' 3' 4'

6' | 5' 6' 7' 8' 1' 2' 3' 4'

7' | 5' 6' 7' 8' 1' 2' 3' 4'

8' | 5' 6' 7' 8' 1' 2' 3' 4'


"Dual":

Denote:

W = {1',2',3',4',5',6',7',8'}, S = {1,2,3,4,5,6,7,8}

where: 1,2,..., are sets of four circles passing through a point
(not be confused with the points 1, 2,... above:
1 above is the common point of the circles 2',3',5',8', while 1 is now the set of the four circles passing through the point 1):

1 = {2',3',5',8'}
2 = {1',3',6',8'}
3 = {1',2',7',8'}
4 = {1',2',3',4'}
5 = {1',4',6',7'}
6 = {2',4',5',7'}
7 = {3',4',5',6'}
8 = {5',6',7',8'}

We have:

1 + 1 = {1',4',6',7'} = 5
1 + 2 = {2',4',5',7'} = 6
1 + 3 = {3',4',5',6'} = 7
1 + 4 = {5',6',7',8'} = 8
1 + 5 = {2',3',5',8'} = 1
1 + 6 = {1',3',6',8'} = 2
1 + 7 = {1',2',7',8'} = 3
1 + 8 = {1',2',3',4'} = 4

etc

Addition Table:

+ | 1 2 3 4 5 6 7 8
-------------------------------
1 | 5 6 7 8 1 2 3 4

2 | 5 6 7 8 1 2 3 4

3 | 5 6 7 8 1 2 3 4

4 | 5 6 7 8 1 2 3 4

5 | 5 6 7 8 1 2 3 4

6 | 5 6 7 8 1 2 3 4

7 | 5 6 7 8 1 2 3 4

8 | 5 6 7 8 1 2 3 4


NOTES:

1. References of the configurations:
Antreas P. Hatzipolakis (et al): Hyacinthos Thread Nice!
Antreas P. Hatzipolakis (et al): Hyacinthos Thread 3+1 Circles

2. Generalization

Παρασκευή, 4 Ιουνίου 2010

 

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