Bakayev E.V.

Inside an isosceles triangle $ABC$, a point $K$ is marked such that $AB = BC = CK$ and $\angle KAC = 30^{\circ}$.

Find the angle $AKB$.
The first solution. Construct an equilateral triangle $ACL$ on $AC$ such that points $L$ and $B$ lie on the same side of $AC$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_7a1b7dea474101c41e1dg-34.jpg?height=538&width=535&top_left_y=11&top_left_x=756)

Draw the altitude $BM$ in triangle $ABC$, which is also the perpendicular bisector of side $AC$. Since $ALC$ is equilateral, point $L$ also lies on line $BM$. Additionally, draw the altitude $AN$ in triangle $ALC$. Since $AN$ is the angle bisector of $\angle LAC$, point $K$ lies on this line. Note also that $K$ lies on the same side of $BM$ as $A$, since due to $CK = CB$, it cannot lie inside triangle $BMC$; thus, $K$ lies on segment $AN$.

Notice that right triangles $BMC$ and $KLC$ are congruent by the leg and hypotenuse ( $MC = AC / 2 = LC / 2 = NC, BC = KC$ ). From this, it follows, first, that $BM = KN$, and second, that $B$ lies on segment $LM$ (since $BM = KN$ and $AB = BC$, thus $\angle ABC > 60^\circ > \angle BAC$. Therefore, point $P$ lies on the other side of line $AC$ from point $B$.

Notice that $\angle BCP = (180^\circ - \angle PBC) / 2$ and $\angle BCA = (180^\circ - \angle ABC) / 2$, since triangles $ABC$ and $PBC$ are isosceles. Then $\angle PCA = \angle PCB - \angle ACB = (180^\circ - \angle PBC) - (180^\circ - \angle ABC) / 2 = (\angle ABC - \angle PBC) / 2 = \angle ABP / 2 = 30^\circ$, since triangle $ABP$ is equilateral. From the equality of angles $\angle KAC = 30^\circ = \angle ACP$, we get $AK \parallel CP$.

The parallelism of $AK$ and $CP$ and the equality $KC = AP$ imply that quadrilateral $AKSR$ is a parallelogram or an isosceles trapezoid. Note that $\angle KAR < \angle BAP = 60^\circ < 90^\circ$ and $\angle KCP < \angle PCB < 90^\circ$ (the angle at the base of an isosceles triangle), but in an isosceles trapezoid, the sum of opposite angles is $180^\circ$, so $AKSR$ is a parallelogram.

It remains to calculate the angles. Let $\angle BAK = \alpha$, then $\angle CAP = \angle BAP - \angle BAK - \angle KAC = 30^\circ - \alpha$.

At the same time, $\angle KCA = \angle CAP$, since $AKSR$ is a parallelogram.

Notice that $\angle BCK = \angle BCA - \angle KCA = \angle BAC - \angle KCA = \alpha + 30^\circ - (30^\circ - \alpha) = 2\alpha$.

Therefore, from the sum of the angles in triangle $BKC$, we get $\angle KBC = (180^\circ - \angle BCK) / 2 = (180^\circ - 2\alpha) / 2 = 90^\circ - \alpha$.

At the same time, from the sum of the angles in triangle $ABC$, we get $\angle ABC = 180^\circ - 2\angle BAC = 180^\circ - 2(30^\circ + \alpha) = 120^\circ - 2\alpha$. Finally, consider triangle $ABK$. In it, the angles $\angle ABK = 30^\circ - \alpha$ and $\angle BAK = \alpha$ are known. Therefore, the third angle $\angle AKB = 150^\circ$.

The third solution. Draw the altitude $BM$ in triangle $ABC$ and denote its intersection with line $AK$ as $O$ (see figure). Notice that $\angle AOM = 60^\circ$ and, by symmetry with respect to $BL$, $\angle COM = 60^\circ$. Then $\angle AOC = 120^\circ$ and $\angle COB = 120^\circ$, that is, rays $OB$ and $OK$ are symmetric with respect to line $CO$. Since points $B$ and $K$ are the intersections of these rays with a certain circle centered at $C$ (which is also symmetric with respect to $CO$), these points are symmetric; in particular, we have $OB = OK$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7a1b7dea474101c41e1dg-35.jpg?height=381&width=626&top_left_y=164&top_left_x=722)

We get that triangle $BKO$ is isosceles with an angle of $120^\circ$ at vertex $O$, from which $\angle OKB = 30^\circ$ and $\angle AKB = 150^\circ$.

## Answer