Not 1789 Sangaku Problem :)

Let $\color{#D61F06}{r}$ be the red sphere’s radius and $\color{#3D99F6}{R}$ the radius of each blue sphere.

We take one section through vertex $E$ and the midpoints $M$ and $N$ of $AB$ and $CD$ respectively (figure 1).

Let $O$ be the center of the red sphere, $F$ , $P$ the contact points of the sphere with the cube and the face $ECD$ , ${E}'$ the midpoint of $MN$ and $c$ the edge of the cube.

Then, triangles $OPE$ and $E{E}'N$ are similar, thus $\frac{OP}{{E}'N}=\frac{OE}{EN}\Leftrightarrow \frac{r}{\frac{s}{2}}=\frac{h-c-r}{\sqrt{{{h}^{2}}+\frac{{{s}^{2}}}{4}}}$ , which solves to $\boxed{\color{#D61F06}{r}=\frac{s\left( h-c \right)}{s+\sqrt{4{{h}^{2}}+{{s}^{2}}}}} \ \ \ \ \ (1)$

Consider the plane $KUV$ that contains the edge $KQ$ of the cube and is perpendicular to $QA$ (hence parallel to $DB$ ) (figure 2). It is easy to see that $KLDU$ is a parallelogram, since $KL\parallel DU$ and $KU\parallel LD$ , thus $AU=s-c$ , $UV=AU\sqrt{2}=\left( s-c \right)\sqrt{2}$ , $KU=LD=KA$ . Hence, $\triangle KUA$ is isosceles with base $UA=s-c$ and height $KT=\sqrt{Q{{T}^{2}}+Q{{K}^{2}}}=\sqrt{A{{T}^{2}}+{{c}^{2}}}=\sqrt{{{\left( \frac{s-c}{2} \right)}^{2}}+{{c}^{2}}}$ .

One blue sphere is inscribed in the tetrahedron $KUAV$ .

We know that for the volume of the tetrahedron holds the relation $V=\frac{1}{3}A\cdot R \ \ \ (2)$ , where $A$ is the total surface area of the tetrahedron and $R$ is the radius of the inscribed sphere.
Now, the volume can also be expressed as $V=\frac{1}{3}\left( \text{area of }\triangle AUV \right)\cdot KQ=\frac{1}{3}\cdot \left( \frac{1}{2}{{\left( s-c \right)}^{2}} \right)\cdot c\Rightarrow V=\frac{1}{6}{{\left( s-c \right)}^{2}}\cdot c \ \ \ \ \ (3)$ .

For the surface area of the tetrahedron we have:
$\begin{aligned} & A=\left( \text{area of }\triangle AUV \right)+\left( \text{area of }\triangle KUV \right)+2\cdot \left( \text{area of }\triangle KUA \right) \\ & =\frac{1}{2}\cdot AU\cdot AV+\frac{1}{2}\cdot UV\cdot KQ+2\cdot \frac{1}{2}\cdot AU\cdot KT \\ & =\frac{1}{2}\cdot {{\left( s-c \right)}^{2}}+\frac{1}{2}\cdot \left( s-c \right)\sqrt{2}\cdot c+2\cdot \frac{1}{2}\cdot \left( s-c \right)\cdot \sqrt{{{\left( \frac{s-c}{2} \right)}^{2}}+{{c}^{2}}}\Rightarrow \\ \end{aligned}$ $A=\frac{1}{2}\cdot \left( s-c \right)\left[ \left( s-c \right)+\sqrt{2}\cdot c+2\cdot \sqrt{{{\left( \frac{s-c}{2} \right)}^{2}}+{{c}^{2}}} \right] \ \ \ \ \ (4)$ Combining $(2)$ , $(3)$ and $(4)$ we get $\boxed{\color{#3D99F6}{R}=\frac{\left( s-c \right)\cdot c}{s-\left( \sqrt{2}-1 \right)c+\sqrt{{{\left( s-c \right)}^{2}}+4{{c}^{2}}}}} \ \ \ \ \ (5)$ .

Since ${\color{#3D99F6}{R}}=2\color{#D61F06}{r}$ , from $(1)$ and $(5)$ we have $\frac{\left( s-c \right)\cdot c}{s-\left( \sqrt{2}-1 \right)c+\sqrt{{{\left( s-c \right)}^{2}}+4{{c}^{2}}}}=2\cdot \frac{s\left( h-c \right)}{s+\sqrt{4{{h}^{2}}+{{s}^{2}}}}$ , which becomes $\frac{\frac{s}{h}-\frac{c}{h}}{\frac{s}{h}+\sqrt{2}+\sqrt{{{\left( \frac{s}{h} \right)}^{2}}+4}}=2\cdot \frac{1-\frac{c}{h}}{1+\sqrt{4{{\left( \frac{h}{s} \right)}^{2}}+1}} \ \ \ \ \ (6)$

In figure 1, triangles $EFK$ and $E{E}'N$ are similar, hence
$\frac{{E}'N}{FK}=\frac{E{E}'}{EF}\Leftrightarrow \frac{\frac{s}{2}}{\frac{c}{2}}=\frac{h}{h-c}\Leftrightarrow c=\frac{hs}{h+s}\Leftrightarrow \begin{cases} \frac{c}{h}=\frac{s}{h+s}=\frac{\frac{s}{h}}{1+\frac{s}{h}} \\ \frac{s}{c}=1+\frac{s}{h} \\ \end{cases}$

By denoting $\frac{s}{h}=x$ , we get $\frac{c}{h}=\frac{x}{1+x}$ and $\frac{s}{c}=1+x$ , thus, $(6)$ consequently becomes $\frac{x-\frac{x}{x+1}}{x+\sqrt{2}+\sqrt{{{x}^{2}}+4}}=2\cdot \frac{1-\frac{x}{x+1}}{1+\sqrt{\frac{4}{{{x}^{2}}}+1}}$ ,
which simplifies to $\left( x-2 \right)\sqrt{{{x}^{2}}+4}=2x+2\sqrt{2}-{{x}^{2}} \ \ \ \ \ (7)$

Squaring both sides, we come to the quadratic $\left( 1+\sqrt{2} \right){{x}^{2}}-2\left( 2+\sqrt{2} \right)x+2=0 \ \ \ \ \ (8)$

The roots of $(8)$ are $x=\sqrt{2}\left( 1\pm \sqrt{2-\sqrt{2}} \right)$ . From these two, $\sqrt{2}\left( 1+ \sqrt{2-\sqrt{2}} \right)$ is the only one that satisfies $(7)$ , hence $\frac{s}{h}=x=\sqrt{2}\left( 1+\sqrt{2-\sqrt{2}} \right)$ .

For the answer, $a=2$ , $b=1$ , $c=2$ , $d=2$ , thus $a+b+c+d=\boxed{7}$ .

Let $s$ be the bottom side of the pyramid, $h$ be the height of the pyarmid, $c$ be the side length of the cube, $r$ be the radius of the small red sphere, and $2r$ be the radius of one of the blue spheres, and label the diagram as follows:

In the top view, $\triangle FPQ$ is a right isosceles triangle where the hypotenuse $FQ = 2r$ , so $QP = FP = \sqrt{2}r$ .

In the side view (where $D$ is directly behind $A$ and $Q$ is directly behind $P$ ), $FP = GH = \sqrt{2}r$ .

By similar triangles, let $\theta = \angle GAF = \angle MAF = \angle JIK = \angle KIN$ . Since $\triangle GAF \sim \triangle JIK$ , $\frac{FG}{AG} = \frac{KJ}{IJ}$ or $\frac{2r}{AG} = \frac{r}{\frac{c}{2}}$ , which solves to $AG = c$ .

On $\triangle IHA$ , $\tan 2\theta = \frac{c}{c + \sqrt{2}r}$ . On $\triangle KJI$ , $\tan \theta = \frac{r}{\frac{c}{2}} = \frac{2r}{c}$ . Since $\tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$ , $\frac{c}{c + \sqrt{2}r} = \frac{2 (\frac{2r}{c})}{1 - (\frac{2r}{c})^2}$ , which solves to $c = (2 - 2\sqrt{2 + \sqrt{2}})r$ , and also means that $\tan 2\theta = \frac{c}{c + \sqrt{2}r} = \frac{2 - 2\sqrt{2 + \sqrt{2}}}{2 - 2\sqrt{2 + \sqrt{2}} + \sqrt{2}}$ .

On $\triangle ELA$ , $\tan 2\theta = \frac{h}{\frac{s}{2}}$ , which rearranges to $\frac{s}{h} = \frac{2}{\tan 2\theta}$ so that $\frac{s}{h} = \frac{2(2 - 2\sqrt{2 + \sqrt{2}} + \sqrt{2})}{2 - 2\sqrt{2 + \sqrt{2}}} = \sqrt{2}(1 + \sqrt{2 - \sqrt{2}})$ . Therefore, $a = 2$ , $b = 1$ , $c = 2$ , $d = 2$ , and $a + b + c + d = \boxed{7}$ .