Problem 11.4. On the side $AC$ of triangle $ABC$, points $M$ and $N$ are marked ($M$ lies on the segment $AN$). It is known that $AB = AN$, $BC = MC$. The circumcircles of triangles $ABM$ and $CBN$ intersect at points $B$ and $K$. How many degrees does the angle $AKC$ measure if $\angle ABC = 68^\circ$?

![](https://cdn.mathpix.com/cropped/2024_05_06_790dd471148872cd0846g-17.jpg?height=379&width=505&top_left_y=1048&top_left_x=468)
Answer: 124.

Solution. From the given in the problem, it follows that $68^{\circ}+\alpha+\gamma=180^{\circ}$, where $\alpha$ and $\gamma$ denote the measures of angles $A$ and $C$ of the triangle, respectively. Since triangle $BAN$ is isosceles, $\angle BNA=90^{\circ}-\frac{1}{2} \alpha$, so $\angle BNC=90^{\circ}+\frac{1}{2} \alpha$. Similarly, $\angle BMA=90^{\circ}+\frac{1}{2} \gamma$.

From the fact that these angles are inscribed in the corresponding circles, we get $\angle BKA=90^{\circ}+\frac{1}{2} \alpha$ and $\angle BKC=90^{\circ}+\frac{1}{2} \gamma$ (Fig. 7). It remains

$$
\angle AKC=360^{\circ}-\angle BKA-\angle BKC=180^{\circ}-\frac{1}{2}(\alpha+\gamma)=90^{\circ}+\frac{1}{2} \cdot 68^{\circ}=124^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_790dd471148872cd0846g-18.jpg?height=414&width=601&top_left_y=92&top_left_x=420)

Fig. 7: to the solution of problem 11.4

Remark. Point $K$ is the incenter of triangle $ABC$.