The important thing first: as usual, this post is

ABSOLUTELY NOT FOR BROADCAST on the Forum !

Now, in the drawing of the last Sangaku in the bottom right corner
there is triangle QLC (75-60-45) which is similar to BPL (75-60-45)

Triangle PQL is right at Q and PL = 2 QL (draw median QW from
Q on PL and WQL is equlateral). So LQC can be rotated 120 deg. into
top portion of PLB with rotation center on angle bisector of PLQ.
Thus incircle of QLC is half of the incircle of BPL.

In right triangle LCD cevian CQ clearly is the cevian of congruent
incircles of LCD and CQD.

Thus we have a pair of obvious 'twin' incircles , and more
hidden ones all over the square, (especially if we do not get
fixated of the numbers 1 and 2 as homothety ratios, we should treat
all numbers equally, egalite tak skazat' ).

No esli ne priniat' specialnyh mer, to eti ... geometry
will go crazy looking for them, odnazhdy uzhe byl pretsedent.

Poetomu, etot poisk dolzhen byt pod surovym kontrolem sverhu.

Salute,

Maj. Paul Litra-Spirta

Hi

> The important thing first: as usual, this post is
>
> ABSOLUTELY NOT FOR BROADCAST on the Forum !

I see this differently: this is extremely important that your message appears publicly. The reason is that a public message can be replied to. And yours definitely warrants a reply.

First, thank you for your insight. I am going to extend that page.

Second, I am humbly asking you to observe that this forum's sole purpose is to discuss mathematics and not getting under each other skin - putting it mildly. I am again asking you to please keep your distance.

> Now, in the drawing of the last Sangaku in the bottom right
>corner
> there is triangle QLC (75-60-45) which is similar to BPL
>(75-60-45)
>
> Triangle PQL is right at Q and PL = 2 QL (draw median QW
>from
> Q on PL and WQL is equlateral). So LQC can be rotated 120
>deg. into
> top portion of PLB with rotation center on angle bisector
>of PLQ.
> Thus incircle of QLC is half of the incircle of BPL.
>
> In right triangle LCD cevian CQ clearly is the cevian of
>congruent
> incircles of LCD and CQD.
>
> Thus we have a pair of obvious 'twin' incircles , and more
> hidden ones all over the square, (especially if we do not
>get
> fixated of the numbers 1 and 2 as homothety ratios, we
>should treat
> all numbers equally, egalite tak skazat' ).

For the sake of those who do not speak Russian, I'll translate the above and the gibberish that followed:

"we should treat all numbers equally, 'egalite' so to speak.

> No esli ne priniat' specialnyh mer, to eti ... geometry
> will go crazy looking for them, odnazhdy uzhe byl
>pretsedent.
>
> Poetomu, etot poisk dolzhen byt pod surovym kontrolem
>sverhu.
>
> Salute,
>
> Maj. Paul Litra-Spirta
>
But, without special actions those ... geometry will go crazy looking for them, a precedent once happened.

For this reason, this search must be carried out under a strict control from above.

Salut,

Maj. Half Litre-(Of alcohol)