In the right-angled triangle $ABC$, the line connecting the midpoint of the leg $CB$ and the center of the inscribed circle intersects the leg $CA$ at point $D$. Express the ratio $AD:DC$ as a function of the angle at $A$!
I. solution. Let the midpoint of side $BC$ be denoted by $F$, and the points of tangency of the incircle with the legs $AC$ and $BC$ by $B_1$ and $A_1$, respectively. We use the fact that in any triangle, with the usual notation, $CB_1 = CA_1 = s - c$, and in our case, since $OA_1CB_1$ is a square, the radius of the incircle is:

$$
\rho = s - c = \frac{a + b - c}{2}
$$

$1986-04-156-1 . \mathrm{eps}$

Figure 1

We will show that the desired point $D$ is the reflection of $B_1$ over the midpoint $G$ of the leg $AC$, or in other words, $DA = B_1C = \rho$. From the obvious similarity of triangles $DOB_1$ and $OFA_1$,

$$
DB_1 : B_1O = OA_1 : A_1F
$$

hence

$$
DB_1 = \frac{\rho^2}{A_1F} = \frac{\rho^2}{\frac{a}{2} - \rho} = \frac{(a + b - c)^2}{2(a - 2\rho)}
$$

We factor the numerator - expanding and using the Pythagorean theorem - as follows:

$$
2c^2 - 2ac - 2bc + 2ab = 2(c - a)(c - b)
$$

and the denominator is equal to $2(c - b)$ by rearranging the second formula for $\rho$. Thus, using $\rho$ again,

$$
DB_1 = \frac{(c - a)(c - b)}{c - b} = c - a = b - 2\rho
$$

therefore $DC = b - \rho$ and $AD = AC - DC = \rho$, as claimed.

Based on this, the ratio to be examined, using the right triangle $AOB_1$,

$$
\frac{AD}{DC} = \frac{B_1O}{AB_1} = \tan \angle B_1AO = \tan \frac{\alpha}{2}
$$

since $AO$ bisects the angle $CAB$.

Remarks. 1. Strictly speaking, the correct expression would be $A_1F = \left|\frac{a}{2} - \rho\right|$, but we immediately realize that the inequality $2\rho > a$ is impossible.

2. It is easy to show that $D$ is the point where the "excircle" tangent to the side $AC$ of triangle $ABC$ touches this side.

This "geometric relationship" obtained through calculation can motivate the search for another solution.

II. solution. Draw the other tangent to the incircle parallel to the leg $AC$. Let this tangent intersect the sides of the triangle at points $A^*$ and $C^*$, and the point of tangency be denoted by $D^*$ (Figure 2).

$1986-04-157-1 . \mathrm{eps}$

Figure 2

The basic idea of the solution - now we are playing with open cards, in the sense of the second remark - is that the triangle $BC^*A^*$, when dilated from $B$, maps to the triangle $BCA$ and $D^*$ maps to $D$. We need to prove this again, but the steps are "self-evident".

We need the midpoint of segment $C^*B$ - let this be $F^*$. Then $FF^* = BF - BF^* = \frac{1}{2}(BC - BC^*) = \frac{1}{2}CC^*$, and this is exactly $\rho$, since $CC^*$ is equal to the diameter of the incircle of triangle $ABC$.

In the quadrilateral $OD^*F^*F$, therefore, $OD^* = FF^* = \rho$, and these two sides are clearly parallel as well. Thus, the quadrilateral is a parallelogram, so the other two sides are also parallel, and this is exactly what we needed, since in the mentioned dilation, $F^*D^*$ maps to the line through $F$ parallel to $F^*D^*$, i.e., to $FD$. Since $A^*C^*$ maps to $AC$, the image of $D^*$ is indeed $D$.

Now we could also fit into the thought process of the previous solution, but based on the above, we can find a quick conclusion.

Due to the found similarity, the ratio $\frac{AD}{DC}$ is equal to $\frac{A^*D^*}{D^*C^*}$. Since $D^*C^* = \rho = D^*O$, the desired ratio is the tangent of $\angle A^*OD^*$ in the right triangle $A^*D^*O$.

Since $\angle A^*OA = 90^\circ - \angle A$ and the angle bisectors of the angles at $A^*$ are parallel - on the one hand, $OD^* \perp AC$ and the angles $\angle A^*OD^*$ and $\angle OAC$ are complementary, and thus equal.

Therefore, the desired ratio is the tangent of the angle $\angle OAC$ - i.e., the tangent of half the angle at $A$.