No, Alex.
The solution should be for general rectangle.
Your case is one special case.
Best regards,
Bui Quang Tuan

The reason is that your file has extension html, which is probably wrong anyway, but, in addition, my site is set up to work with shtml extension. When you try to access a file with extension html, say 123.html, the extention is automatically converted to shtml and the system looks for the 123.shtml file.

HTML file
JPEG picture
ZIP compressed file
Text file
GIF picture
TAR compressed file

First one is format HTML.

The second JPEG format for this particular files with a modest size of 41Kb get rejected for being too large.
Sincerely I believe the instructions deserve an update to level with the excellent content of your site and job.

mpdlc

It is obvious that the circles must be tangent between themselves.

So any increment in size of one radius carried the same decrement in the radius of the other.
Assuming as starting point both have the same radius the resulting area will be A1= 2*pi*R^2

If we allow an increase of d in one circle we will have a decrease of d in the other one so the resulting area will be A2=(pi*(R+d)^2)+ (pi*(R-d)^2)which grows respet to A1 in 2*pi*d^2.

So the strategy is made the first circle as big as possible and fit in the left space the second circle tangent to the first one and to the sides of the rectangle.

>It is obvious that the circles must be tangent between
>themselves.
- What do you think if we allow two circles could be overlapped?

>If we allow an increase of d in one circle we will have a
>decrease of d in the other one
- It is not too clearly that increment and decrement have the same value d.

>So the strategy is made the first circle as big as possible
>and fit in the left space the second circle tangent to the
>first one and to the sides of the rectangle.
- After proofing the strategy is true, we also must construct these two circles by compass and ruler.

Please note that I also still not found the best resolution for this problem.
I am waiting for your nice results.
Best regards,
Bui Quang Tuan

mpdlc

Best regards,
Bui Quang Tuan

Because missing of literatures after more than 2000 years so we don't know and never can know how Archimedes did exactly. I can make only one conjecture of his way. I avoid using function facilities because they discovered later by Newton and Leibniz but I think that Archimedes known about parabola and I use one very simple property of parabola here.

With condition that two circles not overlapping, I think Archimedes did by following way:

1. If L > 2*H, as you already mentioned, he took two congruent circles touching each other and with radius H/2.
2. If H <= L <= 2*H
a). We choose one circle, suppose (Oa) with radius Ra. If this circle is not touching sides of ABCD so we can move this circle to the position in which it touch, say AB, AD then we can make the circle more bigger. So circle (Oa) should be touching AB, AD.
b). We consider to find other circle. This circle must touch (Oa) and can be touching ABCD in three corners B, C, and D. Suppose:
(Ob) touches AB, BC with radius Rb
(Oc) touches BC, CD with radius Rc
(Od) touches CD, DA with radius Rd
and all they are externally touching (Oa).
So we should choose second circle as one from (Ob), (Oc), (Od).
c). We can construct these circles by following way:
- Oa is any point on angle bisector of <DAB with some limited conditions such that it and other circles can be inside ABCD.
- Ob, Oc, Od are on angle bisectors of <ABC, <BCD, <CDA respectively.
- Ob is also on one parabola. It is the locus of all points such that subtraction of distances from it to Oa and to BC is constant = radius of (Oa) = Ra.
- Similarly Od is on one parabola = locus of all points such that subtraction of distances from it to Oa and to CD is constant = Ra.
- Oc is very special case: it is concurrent point of these two parabolas and angle bisector of BCD.
d). So we can construct these four circles and move Oa in its proper range. This range is limited by L, H and can make other circle (if we want to take it as second circle) always inside ABCD.
e). The parabola Ob has one simple property: if Ob is farther from parabola symmetrical axis so radius of (Ob) is bigger. Similarly with parabola Od. Two symmetrical axes of two parabolas are passing through Oa and perpendicular with BC, CD.
Use this property of parabola and range conditions of Oa we can show that: Rb <= Rc and Rd <= Rc. Therefore we should choose (Oa) and (Oc) as our two circles.
f). Now we can use the result in opposite corner case which our friend (Mariano Perez de la Cruz) already proved: choose (Oa) as bigger as possible and (Oc) as smaller as possible (or vice versa). We can calculate:
Ra = H/2, (Oa) touches AB, CD and DA
Rc = H/2 + L - sqrt(2*L*H), (Oc) touches BC and CD and externally touches (Oa)
g). The Oa range:
- Smallest (Oa): when (Ob) = (Oc) and Rb = Rc = H/2, Ra = H/2 + L - sqrt(2*L*H)
- Biggest (Oa): Ra = H/2, Rb = Rc = H/2 + L - sqrt(2*L*H) but in two corners B, C.
h). From expression Rc = H/2 + L - sqrt(2*L*H) we can construct the circle (Oc) by compass and ruler.

I draw this configuration dynamically and see very interesting pictures with exact calculate results. Based on this configuration, we can find other simple calculate proof, but I love this geometrical proof more. Please refer to one picture I attached here.

It is the best conjecture which I can make now. May be we can find other better in the future.

Thank you and best regards,
Bui Quang Tuan

Based on all our discussions I just have found the complete solution for this problem as following:
We consider three right triangles with hypotenuses OaOb, OaOc, OaOd and have:
(Ra + Rb)^2 = (Ra - Rb)^2 + (L - Ra - Rb)^2 (1)
(Ra + Rd)^2 = (Ra - Rd)^2 + (H - Ra - Rd)^2 (2)
(Ra + Rc)^2 = (H - Ra - Rc)^2 + (L - Ra - Rc)^2 (3)
Please note that (3) is in different from with (1), (2)
(1) can be see as quadratic equation of Ra. Solving this equation we have two value of Ra but we can take only one because Ra <= H/2
Ra = (Sqrt(L) - Sqrt(Rb))^2 therefore Ra is min iff Rb is max, i.e. Rb = H/2 and
Ra min = (Sqrt(L) - Sqrt(H/2))^2 (4)
We solve (1), (2), (3) as quadratic equations of Rb, Rd, Rc. After applying conditions of Rb, Rd, Rc we have results:
Rb = (Sqrt(L) - Sqrt(Ra))^2
Rd = (Sqrt(H) - Sqrt(Ra))^2
Rc = (Sqrt(L) - Sqrt(H))^2 - Ra
From these we can calculate:
Rc - Rd = Sqrt(L)*(Sqrt(L) - Sqrt(2*H)) + 2*Sqrt(Ra)*(Sqrt(H) - Sqrt(Ra))
Rc - Rd >= 0 because L>=2*H>0 and H>Ra>0
Rc - Rb = (Sqrt(H) - Sqrt(2*Ra))*(Sqrt(H) + Sqrt(2*Ra) - Sqrt(2*L))
(Sqrt(H) - 2*Sqrt(Ra)) > 0 because H >= 2*Ra
(Sqrt(H) + Sqrt(2*Ra) - Sqrt(2*L)) min iff Ra min and we use (4) for calculation:
(Sqrt(H) + Sqrt(2*Ra) - Sqrt(2*L)) >=
(Sqrt(H) + Sqrt(2*(Sqrt(L) - Sqrt(H/2))^2) - Sqrt(2*L)) =
(Sqrt(H) + Sqrt(2*L) - Sqrt(2*H) - Sqrt(2*L)) = 0
There for Rc - Rb >= 0
At the end we have Rc >= Rb and Rc >= Rd and we must choose two circles (Oa) and (Oc) and can apply Mariano Perez de la Cruz for opposite corner case.
Please check and correct for me if you see any mistake in my calculations

Thank you again for cooperation!
Have you all one nice weekend!
Best regards,
Bui Quang Tuan

Dear Alex and Mariano Perez de la Cruz,

Sorry for one my typo mistake in previous message. The result of Rc by solving equation (3) should be truly read as:
Rc = H + L - Sqrt(2*H*L) - Ra

All the other are still true.

Best regards,
Bui Quang Tuan

https://www.cut-the-knot.org/Curriculum/Geometry/TwoTangentCirclesInPara.shtml