An ancient Japanese geometry problem: Three circles of equal radius inscribed in an isosceles triangle. 
NOTE: This very difficult problem of elementary geometry has an ancient Japanese source (See “Sacred Mathematics: Japanese Temple Geometry”. Princeton University Press, 2008, by F. Hidetoshi & T. Rothman). It was given by F. Hidetoshi to the Spanish international journal “Revista de la O. I. M.” for publication. I was the only one to submit a complete solution, while two Spaniards and a Chilean gave partial solutions. I would love to see another answer but do not know if you can post it here with that precedent.
 A: 
In this diagram,
$$|\overline{AB}| = |\overline{CB}| = a \qquad |\overline{AC}| = 2b \qquad |\overline{AD}| = d \qquad |\overline{CE}| = e$$
Also,
$$x = \frac{|\overline{BD}|}{|\overline{BC}|} \qquad y = \frac{|\overline{AE}|}{|\overline{AD}|} \qquad \angle AEC = 2\theta$$
(where $\angle AEC$ may or may not be a right angle). Writing $T$ for the area of $\triangle ABC$, we see that
$$|\triangle ABD| = Tx \qquad |\triangle CDE| = T(1-x)(1-y) \qquad |\triangle ACE| = T(1-x)y$$

The relation 
$$\text{inradius}\cdot\text{perimeter} = 2\cdot\text{area}$$
implies
$$\begin{align}
r\;\left(\; a+ax+d \;\right) &= 2\;|\triangle ABC| = 2 T x \tag{1}\\
r\;\left(\; a (1-x) + d (1-y) + e \;\right) &= 2\;|\triangle CDE| = 2T(1-x)(1-y) \tag{2}\\
r\;\left(\; e + 2 b + d y \;\right) &= 2\;|\triangle ACE| = 2T (1-x)y \tag{3}
\end{align}$$
We also have
$$T^2 = b^2 ( a^2 - b^2 ) \tag{4}$$
Stewart's Theorem applied to Cevians $d$ and $e$ of $\triangle ABC$ and $\triangle ACD$, ultimately gives us that
$$\begin{align}
d^2 &= a^2 ( 1 - x )^2\phantom{y} + 4 b^2 x \tag{5}\\
e^2 + d^2 y ( 1 - y ) &= a^2 ( 1 - x )^2 y + 4 b^2 ( 1 - y ) \tag{6}
\end{align}$$
Finally, the Law of Cosines provides this relation
$$4 b^2 = d^2 y^2 + e^2 - 2 d e y \cos2\theta \tag{7}$$
Thus, we have seven equations in seven parameters: $a$, $b$, $d$, $e$, $x$, $y$, $T$. There is hope of eliminating six parameters to achieve a relation $F(r,e,\theta) = 0$.

The elimination process is tedious, and I haven't found a particularly "good" path through it. However, I'll note that we can solve $(1)$, $(2)$, $(3)$ that are linear in $a$, $b$, $T$ to get:
$$a = \frac{ex - d(1-y)(1-2x)}{(1 - x)(1 - y - x y )}
\qquad b = \frac{2 e (1+x)y - e - d (1-2x)y}{2(1 - y - x y)} \tag{$\star$}$$
$$T = r\;\frac{e (1+x) + 2 d x - d (1+x) y}{2( 1 - x)( 1 - y - x y)}$$
This immediately reduces our burden a bit. Also, we can solve $(5)$ and $(6)$ for $a^2$ and $b^2$
$$a^2 =\frac{d^2 (1-y-xy+xy^2) - e^2 x}{(1 - x)^2(1 - y - x y)} \qquad 
b^2 = \frac{e^2 - d^2 y^2}{4(1 - y - x y)} \tag{$\star\star$}$$
which helps in quickly establishing further reductions. (For instance, one can equation $a^2$ from $(\star)$ with $a^2$ from $(\star\star)$.)
With the help of Mathematica, I was able to wade through a river of resultants to get to this relation

$$\begin{align}  
\phantom{\cdot}
&\left(\;2 \hat{r}^3 + \hat{r}^2 \hat{e}\;(1-3\sin^2\theta) - 2 \hat{r} \hat{e}^2 \sin^2\theta + \hat{e}^3 \sin^4\theta\;\right) \\
\cdot 
&\left(\;2 \hat{r}^3 - \hat{r}^2 \hat{e}\;(1-3\sin^2\theta)- 2 \hat{r} \hat{e}^2 \sin^2\theta - \hat{e}^3 \sin^4\theta\;\right) = 0
\end{align}$$

where $\hat{r} := r\sin\theta$ and $\hat{e} := e\cos\theta$.
(Of course, since the factors are merely cubics, we can solve for roots $\hat{r}$ explicitly. This is left as an exercise for the reader.)
When $\angle AEC$ is a right angle, $\theta = \pi/4$ and the equation reduces to 

$$(4r - e) (2r^2 - e^2)\cdot(4r + e) (2r^2 - e^2) = 0$$

Since $r$ and $e$ are non-negative (and $r \leq e/2$), we conclude $r = e/4$.
A: J. Marshall Unger (Department of East Asian Languages and Literatures, Ohio State University) has a new book:
Sangaku Proofs: A Japanese Mathematician at Work
(January, 2015, Cornell East Asia)
Elsewhere, Unger has discussed other Sangaku problems, including this one.  Unger analyzes Kitagawa's proof, noting certain unjustified leaps in the reasoning.  Then he gives a new elementary proof of his own.  (Cut-The-Knot LINK for Unger's solution.)
A: Oh! You can see my solution here: http://www.oei.es/oim/revistaoim/numero35/165.pdf 
Clearly, If this solution is correct which I hope and believe, one has F(X, r) = (X -4r)^2 where F(X, r) = 0 is my answer. But I did not even think about this desirable simplification (work to go to F was quite hard).
A: 
Given the isosceles triangle $ABC$
with unit legs $|AC|=|BC|=1$ 
and the base $|AB|=c\in(0,2)$,
$|AD|=v$, $|BE|=u$, $|BD|=t$,
inscribed circles 
of $\triangle ADC$, $\triangle ABE$
and $\triangle BDE$
has the same radius $r_c$.
Obviously, the value of $c$
uniquely defines 
all the other parameters of this sangaku configuration,
that is, the corresponding values of $t,v,u$ and $r_c$,
but to find an explicit expressions for 
$t(c),v(c),u(c)$ and $r_c(c)$ is not an easy task.
However, $v,u,r_c$
can be easily expressed 
in terms of $c$ and $t$:
\begin{align} 
v^2&=t^2+c^2(1-t)
\tag{1}\label{1}
,\\
4u^2 &= t c(2 +c)
\tag{2}\label{2}
,\\
r_c &= \frac{c(1-t)\sqrt{4-c^2}}{2\Big(2-t+\sqrt{t^2+c^2-t\,c^2}\Big)}
\tag{3}\label{3}
.
\end{align}
Using the two other questions,
Is there a way to reduce a specific quintic to cubic?
and
Approximating function for the root of quintic polynomial,
the relationship between $r$ and $u$
can be represented 
by a quarter of a nice 
$\infty$-shaped curve:

Interestingly, the solution $u=4r_c$ to the original sangaku
construction, when $\angle DEB=90^\circ$
and hence $c=t=0.56$, is not unique.
There is another configuration, when $u=4r_c$:
\begin{align} 
c & \approx 0.13935311959635
,\\
t & \approx 0.38561654555749
,\\
u&  \approx 0.16953033464425
,\\
r_c&\approx 0.04238258366106
,\\
\angle DEB&\approx 128^\circ
.
\end{align} 
