- Dear Paul,

You wrote:>Given triangle ABC, it is easy to construct the two circles,

Just to small comments:

>each passing through A, and tangent to BC at one of the vertices B

>and C.

>

>How does one construct the circle tangent to BC, and to each of these

>two circles externally?

>

>There is of course a construction by inversion. I have found a very

>simple construction of this circle, and verified the correctness by

>calculations. I wonder if the following construction is known and if

>one can find a synthetic proof:

>

>Let the bisector of angle A intersect BC at P. Construct the

>perpendicular to BC at P to intersect the median AD at Q.

>The cirlce with PQ as diameter is tangent BC and to each of the two

>circles above.

Since the circles passing through A and tangent to BC at B,C, resp.,

have another intersection A', means that the same construction should

apply for the triangle A'BC.

Since there are two solutions of the problem (ie the circle in question

touches BC between B, C or at a point on its extension), means that there

should be a similar construction for the second circle.

(in this case the bisector should be the bisector of the external angle A).

Antreas - Dear Paul,

You wrote:>Given triangle ABC, it is easy to construct the two circles,

Here is my proof:

>each passing through A, and tangent to BC at one of the vertices B

>and C.

>

>How does one construct the circle tangent to BC, and to each of these

>two circles externally?

>

>There is of course a construction by inversion. I have found a very

>simple construction of this circle, and verified the correctness by

>calculations. I wonder if the following construction is known and if

>one can find a synthetic proof:

>

>Let the bisector of angle A intersect BC at P. Construct the

>perpendicular to BC at P to intersect the median AD at Q.

>The cirlce with PQ as diameter is tangent BC and to each of the two

>circles above.

Assume that AC > AB or b > c

A

/\

/ \

/ \

/ \

/ \ K

/ M |

/ \ |

/ Q \ |

/ \ |

/ N-------------L

/ | \|

B-----H---P--D---------|C

AP: int. bisector of ang(BAC) = A

AD: a-median (D = midpoint of BC)

AH: a-altitude

M: midpoint of AC

Q = AD /\ perp. to BC at P

N = Midpoint of QP = center of the circle with diameter QP

K = (perp. to AC at M) /\ (perp. to BC at C) : it is the center of the circle

passing through A and tangent to BC at C (KC = radius of this circle)

L= orth. projection of N on KC.

PQ PD AH * PD

Tr.(HAD) ~ Tr.(PQD) ==> ---- = ---- ==> PQ = -------

AH DH DH

We have:

PD = PC - DC, PC = ab/(b+c), DC = a/2, DH = (b^2 - c^2)/2a

So, we get: PQ = AH * a^2 / (b+c)^2.

To prove that the circle (N, PQ/2) is tangent externally to circle (K, KC)

is enough to prove that: KN = radius of the one circle + rad. of the other

circle

that is: KN = KC + NP or

KN^2 + KC^2 + NP^2 + 2KC*NP (1)

We have: KN^2 = KL^2 + LN^2 = (KC-LC)^2 + LN^2 = (KC - NP)^2 + PC^2 =

KN^2 = KC^2 + NP^2 - 2KC*NP + PC^2 (2)

From (1) and (2) we get that it is enough to prove that PC^2 = 4KC*NP (3).

Is this last equality true? Let's see:

We have:

PC^2 = a^2b^2 / (b+c)^2, NP = PQ/2 = (AH*a^2)/2(b+c)^2

KC MC KC b

Tr. (MKC) ~ Tr.(HCA) ==> ---- = ---- or ---- = ---- ==> KC = b^2 / 2AH

AC AH b 2AH

So, the (3) becomes: a^2b^2 / (b+c)^2 = 4 * (b^2 / 2AH) * [(AH*a^2)/2(b+c)^2]

==> 1 = 1. True!

Similarly we prove that the circle (N, PQ/2) touches externally the circle

passing through A and tangent to BC at B.

Antreas - [PY]:
>>Given triangle ABC, it is easy to construct the two circles,

[I changed the notation of the points. Similarly we define the points

>>each passing through A, and tangent to BC at one of the vertices B

>>and C.

>>

>>How does one construct the circle tangent to BC, and to each of these

>>two circles externally?

>>

>>There is of course a construction by inversion. I have found a very

>>simple construction of this circle, and verified the correctness by

>>calculations. I wonder if the following construction is known and if

>>one can find a synthetic proof:

>>

>>Let the bisector of angle A intersect BC at A'. Construct the

>>perpendicular to BC at A' to intersect the median AA" at Pa.

>>The cirlce with A'Pa as diameter is tangent BC and to each of the two

>>circles above.

B', C'; B", C"; Pb, Pc]

1. Let Ma, Mb, Mc be the centers of Paul's circles (ie midpoints

of A'Pa, B'Pb, C'Pc).

Are the lines AMa, BMb, CMc concurrent?

2. Let P'a, P'b, P'c be the reflections of Pa, Pb, Pc in BC, CA, AB, resp.

Are the lines AP'a, BP'b, CP'c concurrent?

3. Let M'a, M'b, M'c be the reflections of Ma, Mb, Mc on BC, CA, AB, resp.

Are the lines AM'a, BM'b, CM'c concurrent?

Antreas

PS: A typo in my proof:

> KN^2 + KC^2 + NP^2 + 2KC*NP (1)

^

KN^2 = KC^2 + NP^2 + 2KC*NP (1)

aph