## A simple construction

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• Dear friends, Given triangle ABC, it is easy to construct the two circles, each passing through A, and tangent to BC at one of the vertices B and C. How does
Message 1 of 2 , Mar 31, 2001
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Dear friends,

Given triangle ABC, it is easy to construct the two circles,
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.

Best regards
Sincerely,
Paul
• Paul recently asked for a synthetic proof of a neat construction ... ... He mentioned a construction by inversion and I wondered whether this would not also
Message 2 of 2 , Apr 2, 2001
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Paul recently asked for a synthetic proof of a neat construction ...

> Given triangle ABC, it is easy to construct the two circles,
>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?

>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.

He mentioned a construction by inversion and I wondered whether this would
not also prove his direct one. So I start with what is going to be an
inverse figure ....

Consider a triangle ABC with internal bisector at A meeting the
circumcircle S at P and with tangents at B,C meeting at K. A circle S1
touches these tangents and the circle S at P. KA meets the circles S,S1
in corresponding points A,A1 and Q,Q1; KP meets S1 again at P1. Note
angles QAP=Q1A1P1=Q1PP1 so that the circle Q1PA touches KP at P, ie is
orthogonal to the circle S.

Inverting w.r.t. A proves Paul's construction (with angles B,C
interchanged). This invokes the fact that the adjoint circles through A
(touching BC at B, C respectively) meet on the median from A (in the above
case this intersection is the inverse of K)

Dick Tahta
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