
Zadatak B-4.2. Neka je $A B C$ jednakostraničan trokut. Na pravcu $A B$ odabrana je točka $P$ tako da točka $A$ leži između točke $P$ i točke $B$. Neka je $a$ duljina stranice trokuta $A B C, r_{1}$ polumjer upisane kružnice trokutu $P A C$ i $r_{2}$ polumjer kružnice pripisane trokutu $P B C$ u odnosu na stranicu $B C$. Odredite zbroj $r_{1}+r_{2}$ kao funkciju od $a$.


## Rješenje.

![](https://cdn.mathpix.com/cropped/2024_05_30_c2de659189058fc448a1g-17.jpg?height=591&width=1065&top_left_y=1857&top_left_x=519)
$\angle T_{1} O_{1} R=60^{\circ}$ (suplement kuta $\angle T_{1} A R=120^{\circ}$ )

$\angle A O_{1} T_{1}=30^{\circ}$

$$
\begin{gathered}
\left|T_{1} T_{2}\right|=\left|T_{1} A\right|+|A B|+\left|B T_{2}\right| \\
\operatorname{tg} 30^{\circ}=\frac{\left|T_{1} A\right|}{r_{1}}, \quad\left|T_{1} A\right|=r_{1} \cdot \operatorname{tg} 30^{\circ}=\frac{r_{1}}{\sqrt{3}}
\end{gathered}
$$

Analogno, $\angle B O_{2} T_{2}=30^{\circ}$

$$
\operatorname{tg} 30^{\circ}=\frac{\left|B T_{2}\right|}{r_{2}}, \quad\left|B T_{2}\right|=r_{2} \cdot \operatorname{tg} 30^{\circ}=\frac{r_{2}}{\sqrt{3}}
$$

Uvrstimo (2) i (3) u (1).

$$
\begin{aligned}
& \left|T_{1} T_{2}\right|=\frac{r_{1}}{\sqrt{3}}+a+\frac{r_{2}}{\sqrt{3}}=\frac{r_{1}+r_{2}}{\sqrt{3}}+a \\
& \left|T_{1}^{\prime} T_{2}^{\prime}\right|=\left|T_{1}^{\prime} C\right|+\left|C T_{2}^{\prime}\right|=|C R|+|C S|=(a-|R A|)+(a-|B S|) \\
& |A R|=\left|T_{1} A\right| \quad \text { i } \quad|B S|=\left|B T_{2}\right| \\
& \left|T_{1}^{\prime} T_{2}^{\prime}\right|=2 a-\frac{r_{1}+r_{2}}{\sqrt{3}}
\end{aligned}
$$

Kako je $\left|T_{1} T_{2}\right|=\left|T_{1}^{\prime} T_{2}^{\prime}\right|$, imamo:

$$
\begin{aligned}
& 2 a-\frac{r_{1}+r_{2}}{\sqrt{3}}=\frac{r_{1}+r_{2}}{\sqrt{3}}+a \\
& 2 \cdot \frac{r_{1}+r_{2}}{\sqrt{3}}=a \\
& r_{1}+r_{2}=\frac{a \sqrt{3}}{2}
\end{aligned}
$$

