$\left.\begin{array}{l}\text { [ Properties of angle bisectors, concurrency ]} \\ \text { [Two tangents drawn from the same point]}\end{array}\right]$

In triangle $K L M$, the angle bisector $K P$ is drawn. The incircle of triangle $K L P$ touches side $K L$ at point $Q$, and $L Q=a$. Points $E$ and $R$ are chosen on sides $K L$ and $L M$ respectively such that line $E R$ passes through the incenter of triangle $K L M$. Find the length of the angle bisector $K P$, given that $E L + L R = b$, and the ratio of the areas of triangles $K L P$ and $E L R$ is $\alpha$.
## First method.

Let $O$ be the center of the circle of radius $r$ inscribed in triangle $KLM$, and $A$ and $B$ be the points of tangency with sides $LM$ and $KL$ respectively. Then

$$
S_{\triangle KLP} = \frac{1}{2} LP \cdot OA + \frac{1}{2} KL \cdot OB = \frac{1}{2}(LP + KL) r,
$$

$$
S_{\triangle ELR} = \frac{1}{2} LR \cdot OA + \frac{1}{2} LE \cdot OB = \frac{1}{2}(LR + LE) r = \frac{1}{2} br,
$$

and since

$$
\frac{S_{\Delta KLP}}{S_{\triangle ELR}} = \frac{LP + KL}{b} = \alpha
$$

we have $LP + KL = \alpha b$.

Let $p$ be the semiperimeter of triangle $KLP$. The circle inscribed in triangle $KLP$ touches side $KP$ at point $Q$, so

$$
a = LQ = p - KP = \frac{LP + KL - KP}{2} = \frac{\alpha b - KP}{2}
$$

from which we find that $KP = \alpha b - 2a$.

## Second method.

Let $O$ be the center of the circle inscribed in triangle $KLM$. Then $LO$ is the angle bisector of triangles $KLP$ and $ELR$. Denote $\angle KLM = \gamma$. By the formula for the angle bisector of a triangle,

$$
LO = \frac{2 \cdot LE \cdot LR \cdot \cos \frac{\gamma}{2}}{LE + LR} = \frac{2 \cdot LK \cdot LP \cdot \cos \frac{\gamma}{2}}{LK + LP}
$$

Therefore,

$$
\frac{LK \cdot LP}{LK + LP} = \frac{LE \cdot LR}{LE + LR} = \frac{LE \cdot LR}{b} \Rightarrow LK + LP = b \cdot \frac{LK \cdot LP}{LE \cdot LR}
$$

and since

$$
\frac{S_{\triangle KLP}}{S_{\triangle ELR}} = \frac{\frac{1}{2} \cdot LK \cdot LP \cdot \sin \gamma}{\frac{1}{2} \cdot LE \cdot LR \cdot \sin \gamma} = \frac{LK \cdot LP}{LE \cdot LR} = \alpha
$$

we have $LK + LP = \alpha b$.

From the equality of the segments of tangents drawn from a point to a circle, it follows that $KP = LK + LP - 2$.

$LQ = \alpha b - 2a$.

## Answer

$\alpha b - 2a$.