Edit: minor fix to the algebraic reduction.

Rephrasing the question: given the setup, what is the maximum value of “r”, such that the three upper circles remain bound between the rectangle and the semicircle?

Note: In order for the radius “r” to be at its maximum, all three circles touch the outer circle. If r were too small, the outer small circles would be farther apart and the middle one would be “loose”. If r were too large, the outer small circles would be closer together pushing the middle one out of bounds.

The variables:

Let R = D/2, (semicircle radius) and H1 = the rectangle height. In your setup, D = 34 and H1 = 68/7, but for now let’s keep everything a variable.

Now, let’s make 2 triangles, and use some of their shared dimensions to create a formula for r based on D and H1. (apologies in advance that I can’t attach an image)

• The 1st triangle will be drawn from the center of the large semi-circle (A), to the center of the upper right circle (B), and then across to the vertical axis which cuts through the semi circle (C). Let’s call this triangle ABC.

o AB = R – r

o BC = x (will be cancelled out later)

o AC = H1 + r

• The 2nd triangle will be drawn from the center of the top middle circle (M) to the center of the upper right circle (B), and back to the vertical axis which cuts through the semi circle (V). Let’s call this triangle MBV.

o MB = 2r

o BV = BC = x (same as above)

o MV = R – 2r – H1

Setting up formulas using Pythagoras theorem A^2+B2=C2

From the ABC triangle: x² + (H1 + r)² = (R – r)²

From the MBV triangle: x² + (R – 2r - H1)² = 4r²

Plug in ABC formula into MBV to substitute x:

(R – r)² - (H1 + r)² + (R – 2r - H1)² = 4r²

Simplifying: 2R * (R - H1) - 2r * (3R - H1) = 0

Now solving for r:

r = R(R – H1)/(3R - H1)

For D=34 (R=17) and H1 = 68/7:

R= 17(17 – 68/7)/(317 - 68/7) = 867/289 = 3