Dear Hyacinthists,

I wrote

> There is a little problem : there are 4 conics going through 3

given

> points and touching 2 given lines.

> Of course, these 4 conics are constructible.

Here is a possible - and quite classical - construction of the conics

going through A,B,C and touching the lines L1 and L2. (at most 4 real

solutions)

C1,C2 = common points of AB with L1,L2

B1,B2 = common points of AC with L1,L2

The radical axis of the circles (O) = ABC and C1C2C intersects AB at

Oc; the circle centered at Oc and orthogonal to (O) intersects AB at

P,P'.

Same way, the radical axis of the circles (O) and B1B2B intersects AC

at Ob; the circle centered at Ob and orthogonal to (O) intersects AC

at Q,Q'.

Now take L = PQ or PQ' or P'Q or P'Q'

If L intersects L1 and L2 at D1 and D2, the conic ABCD1D2 is a

solution

This due to the fact that

- P, P' are the fixed points of the involution on the line AB

swapping (A,B) and (C1,C2)

- Q, Q' are the fixed points of the involution on the line AC

swapping (A,C) and (B1,B2)

- As the line D1D2 (twice) is member of the pencil generated by the

required conic and the degenerated conic L1 union L2, the common

point of D1D2 and AB - or AC - must be a fixed point of the

involution i - ot i' -

Friendly. Jean-Pierre