Given a triangle $ABC$ with the base $a$ and the sum of the other two sides $b+c$. Determine this triangle when the median and the angle bisector from $A$ form the largest possible angle with each other.
Let the midpoint of side $BC$ be $D$, and the angle in question be $x$. Then:

$$
\frac{a}{2}: b=\sin \left(\frac{\alpha}{2}+x\right): \sin C D A
$$

and

$$
\frac{a}{2}: c=\sin \left(\frac{\alpha}{2}-x\right): \sin C D A
$$

from which

$$
\frac{b}{c}=\frac{\sin \left(\frac{\alpha}{2}-x\right)}{\sin \left(\frac{\alpha}{2}+x\right)}
$$

or

$$
\frac{b}{c}=\frac{\sin \frac{\alpha}{2} \cos x-\cos \frac{\alpha}{2} \sin x}{\sin \frac{\alpha}{2} \cos x+\cos \frac{\alpha}{2} \sin x}=\frac{1-\operatorname{ctg} \frac{\alpha}{2} \operatorname{tg} x}{1+\operatorname{ctg} \frac{\alpha}{2} \operatorname{tg} x}
$$

From this,

$$
\operatorname{tg} x=\frac{c-b}{(b+c) \operatorname{ctg} \frac{\alpha}{2}}
$$

or

$$
\operatorname{tg} x=\frac{(c-b) \sqrt{(s-b)(s-c)}}{(c+b) \sqrt{s(s-a)}}
$$

$b+c, b+c+a, \frac{b+c}{2}-\frac{a}{2}$ are constants, so for $\operatorname{tg} x$ and thus $x$ to be as large as possible, the numerator must be maximized, that is, this function:

$$
(c-b)^{2} \frac{a+c-b}{2} \times \frac{a+b-c}{2}=y^{2}
$$

or

$$
(c-b)^{2} a^{2}-(c-b)^{4}=4 y^{2}
$$

The maximum is achieved when

$$
c-b=\frac{a}{2} \sqrt{2}
$$

but

$$
c+b=c+b
$$

so

$$
c=\frac{a \sqrt{2}+2(b+c)}{4} ; b=\frac{-a \sqrt{2}+2(b+c)}{4}
$$

(István Szabó.)