$\left.\begin{array}{l}\text { [ Homothety helps solve the problem. } \\ \text { [Two tangents drawn from one point] }\end{array}\right]$

On the sides $A C$ and $B C$ of triangle $A B C$, points $P$ and $Q$ are marked, respectively. It turned out that $A B=A P=B Q=1$, and the point of intersection of segments $A Q$ and $B P$ lies on the inscribed circle of triangle $A B C$. Find the perimeter of triangle $A B C$.
Let the incircle of triangle $ABC$ touch its sides $AB, BC$, and $AC$ at points $S, R$, and $T$; $M$ is the intersection point of segments $AQ$ and $BP$. We can observe that point $P$ lies on segment $CT$, and point $Q$ lies on segment $CR$. Then segments $AM$ and $BM$ intersect the circle again. Let the points of intersection be $K$ and $L$ respectively. If $I$ is the center of the circle, then $BI$ is the angle bisector of $\angle BAC$. Under symmetry with respect to line $BI$, point $S$ maps to $R$, point $A$ maps to $Q$ (since $BQ = BA$), and the circle maps to itself. Therefore, arcs $KS$ and $MR$ are equal. Similarly, we can prove that arcs $SL$ and $MT$ are equal. Since $BI \perp AM$ and $AI \perp BM$, $T$ is the orthocenter of triangle $ABM$. Therefore, $MI \perp AB$, and $MS$ is the diameter of the circle. Hence, arcs $KT$ and $RL$ are equal. Consequently, arc $KTMR$ is $180^\circ$, meaning $KR$ is the diameter of the circle. Under a homothety centered at $A$, which maps the incircle to the excircle of triangle $ABC$, the tangent $l$ at point $K$ to the incircle will map to a parallel tangent to the excircle, i.e., to line $BC$, and ray $AK$ will map to itself. Therefore, $Q$ is the point of tangency of the excircle with side $BC$. Then, if $p$ is the semiperimeter of triangle $ABC$, we have

$$
1 = BQ = p - AB = p - 1
$$

From this, we find that $p = 2$. Therefore, the perimeter of triangle $ABC$ is 4.

![](https://cdn.mathpix.com/cropped/2024_05_06_8a44229da56811dd86d6g-25.jpg?height=871&width=686&top_left_y=-1&top_left_x=709)

## Answer

4.00