Proportions of ornamental gables in gothic architecture I'm trying to figure out mathematically precise proportions for gothic architecture-style gables.
The gable has the shape of a triangle. There is a central incircle, as well as a smaller incircle in the remaining space at the top. The legs of the triangle are tangent to two additional circles of the same radius as the smaller circle at the top, which are also tangent to the central incircle. If the base of the triangle is drawn so that it is tangent to the central incircle, then the two neighboring circles at the bottom are cut exactly in half.

I hope the image makes it more clear.
Now, my question is: for any given base or height of the triangle, or for any given radius of the central incircle, what is the radius of the smaller circles?
Since the legs of the triangle are tangent to 3 circles on either side, which also touch, and since the smaller circles are of the same size, I feel like there should be exactly one answer where the proportions exactly match.
I've tried a few things but my maths is a bit rusty and I can't seem to make a lot of progress. :/ I'd be grateful even for a suggestion from which direction to tackle this problem.
 A: 
Denote the radius $R$ and $r$, all sides are calculated using similar triangles and Pythagoras theorem.
On the big triangle similarity we obtain
$${{2r\sqrt{Rr}\over R-r}+2\sqrt{Rr}+2\sqrt{Rr}+{r(R-r)\over 2\sqrt{Rr}}\over {r(R+r)\over 2\sqrt{Rr}}+\sqrt{r^2+2Rr}}={R+r\over R-r}$$
Simplify,
$${{2r^2R\over R-r}+4Rr+{r(R-r)\over 2}\over {r(R+r)\over 2}+r\sqrt{Rr+2R^2}}={R+r\over R-r}$$
I guess doing the algebra will indeed simplify and be factored this as there is a nice integral ratio and two ratios that looks like solution to a quadratic equation, but I just used wolfram alpha to get the three possible ratios:
$$R={1\over 7}r$$
$$R={5+\sqrt{17}\over 4}r$$
$$R={5-\sqrt{17}\over 4}r$$
Out of the three possibilities, only one is larger than $1$ so
$$R={5+\sqrt{17}\over 4}r$$
is the answer.
Appendix: what I entered into wolfram-alpha
(2x^2y/(y-x)+4xy+x(y-x)/2)(y-x)=(x(y+x)/2+xsqrt(yx+2y^2))(y+x)

A: 
We have isosceles $\triangle ABC$ with $AB = AC$.  $O$ is the midpoint of $BC$, and angle $ t = \angle OAC $.  Further the height is $h = AO$
From this, it follows that
$OB = OC = h \ \tan t $
The semi-perimeter is given by
$ s = OB + AB = h ( \tan t + \sec t ) $
The radius of the incircle (shown in red) is
$ R = \dfrac{\text{Area}}{s}  = \dfrac{h \tan t }{\tan t + \sec t } $
Let $f = \dfrac{1 - \sin t }{1 + \sin t } $, then the radius of the small circle on top is $r$ which is given by
$ r = f R $
This is also the radius of each of the small circles on either side on the bottom.  Since their centers is on the base, then the distance of the center of the bottom right circle $G$ from the base midpoint $O$ is
$ d = h \tan t - r \sec t = h ( \tan t - f \dfrac{\tan t }{\tan t + \sec t } )$
Applying Pythagorean theorem to $\triangle ONB $
$ d^2 + R^2 = (R + r)^2 $
which simplifes to
$ d^2 = 2 R r + r^2 = (f^2 + 2 f ) R^2$
Substituting $d$ and $R$, and dividing by $h^2$ gives us
$ ( \tan t - f \dfrac{\tan t }{\tan t + \sec t } )^2 = (f^2 + 2 f) \dfrac{\tan^2 t }{(\tan t + \sec t )^2} $
Multiplying through by $ (\tan t + \sec t )^2 $, gives
$ ( \tan t (\tan t + \sec t ) - f \tan t )^2 = (f^2 + 2 f) \tan^2 t $
Dividing through by $\tan^2 t $, gives
$ ( \tan t + \sec t - f)^2 = f^2 + 2 f $
Substituting $f$ and multiplying through by $(1 + \sin t )^2 $, gives us
$ ( (\tan t  + \sec t) (1 + \sin t ) - (1 - \sin t )  )^2 = (1 - \sin t )^2 + 2 (1 - \sin t)(1 + \sin t) $
Multiplying through by $ \cos^2 t $, simplifies the above expression to,
$ ( (\sin t + 1)^2 - 1 + \sin t)^2 = (1 - \sin^2 t) ( 3 - 2 \sin t - \sin^2 t )$
Simplifying,
$ ( \sin^2 t + 3 \sin t)^2 = \sin^4 t + 2 \sin^3 t - 4 \sin^2 t - 2 \sin t + 3 $
And further,
$ \sin^4 t + 6 \sin^3 t + 9 \sin^2 t =\sin^4 t + 2 \sin^3 t - 4 \sin^2 t - 2 \sin t + 3 $
And finally,
$ 4 \sin^3 t + 13 \sin^2 t + 2 \sin t - 3 = 0 $
which is a cubic polynomial equation in $\sin t$.  Feeding this to wolframalpha.com gives only one valid solution which is
$ \sin t = \dfrac{\sqrt{17} - 1} { 8 } $
This is the golden trigonometric ratio for this problem.  It follows that
$t= \sin^{-1} \bigg(\dfrac{\sqrt{17} - 1} { 8 }\bigg) \approx 22.9786^\circ $
So, given $h$ we can now compute all the required radii using this value of $t$.
Finally, note the ratio
$\dfrac{r}{R} = f = \dfrac{1 - \sin t }{1 + \sin t } = \dfrac{ 9 -\sqrt{17} }{ \sqrt{17} + 7 } = \dfrac{ (9 - \sqrt{17})(7 - \sqrt{17}) }{49 - 17} \\= \dfrac{ 80 - 16 \sqrt{17} }{32} = \dfrac{5 - \sqrt{17}}{2} $
The inverse ratio is
$ \dfrac{R}{r} = \dfrac{2}{5 - \sqrt{17}} = \dfrac{ 2 (5 + \sqrt{17} ) }{ 8 } = \dfrac{5 + \sqrt{17}}{4} $
A: 
I propose to recast your issue into a different context.
Consider the image above featuring an isosceles triangle $ABC$ with $P_1$ the foot of its main height, its inscribed circle (radius $R$) and two little circles (radii $r$) inscribed into basis angles $A$ and $B$ and tangent to the inscribed circle.
What is the connection with the initial figure ? If the ratio $R/r$ is well chosen, this figure can be placed in exact coincidence with a tilted version (say around 110°) of the initial figure where the former horizontal line has become the red line ; please note that the second little circle "cutted in half" isn't present. In fact, the symmetry of the figure with respect to line bissector $AC_1$ warrants that working on the circle with center $C_2$ is sufficient.
Let $2 \alpha$ be the angle in $A$. As illustrated in the figure, angle $\alpha$ is found at different places. Let $c:=\cos \alpha$ and $s:=\sin \alpha$. Taking the horizontal and vertical axes intersecting in $P_1$ as the axes of coordinates, the coordinates of $C_2$ are :
$$C_2: \ (C_1C_2 \cos \alpha, DP_1) \ = \ ((R+r)c,R-r)\tag{1}$$
Besides, in triangle $C_1DC_2$, we have :
$$\tan \alpha = \frac{DC_1}{DC_2}=\frac{(R-r)}{(R+r) \cos \alpha} $$
As a consequence : $$s:=\sin \alpha=\frac{(R-r)}{(R+r)}\tag{2} $$
Now, let us consider the line $(L)$ issued from point $C_2$ orthogonal to line $AC_1$ (red line in the picture) : this is the horizontal line we had in the initial figure.
The equation of (L), due to the fact that its slope is $\alpha - \pi/2$, and to the coordinates of $C_2$ is :
$$y - r = -\frac{1}{\tan \alpha}(x- (R+r)c)$$
Otherwise said, taking into account the definition $\tan \alpha =\frac{s}{c}$:
$$s(y-r)=-c(x-(R+r)c) \ \iff \ cx+sy-rs-(R+r)c^2=0$$
Now the constraint is that dist$(C_1,(L)) = R$
The formula for the distance from a point with coordinates $(x_0,y_0)$ to a line with equation $ax+by+c+0$ is known to be
$$\frac{|a[x_0]+b[y_0]+c|}{\sqrt{a^2+b^2}}$$
giving in our case the following form to our constraint (remember that the coordinates of $C_1$ are $(0,R)$) :
$$|c[0]+s[R]-rs-(R+r)c^2|=R \ \iff \ (R+r)c^2 - s(R-r) = R\tag{3}$$
(we have removed the absolute value by replacing its contents by its opposite ; in fact, a little analysis shows that it must be so).
Plugging the expression (2) of $s$ into (3), we get the following constraint :
$$2R^2-5Rr+r^2=0$$
which (divided by $r^2$) gives a quadratic equation in variable $\rho=R/r$ whose unique valid solution ($\rho$ has to be $>1$) is
$$\rho = \frac{R}{r} = \frac{5+\sqrt{17}}{4} \approx 2.28 \tag{4}$$
Plugging this value of $\rho$ into relationship (2) gives:
$$\sin \alpha = \frac{\rho -1}{\rho+1}= \frac{-1+\sqrt{17}}{8} \  \implies \ \alpha \approx 22.98°$$
