In triangle $ABC$, $\angle BAC = 45^{\circ}$. Let $P$ be the point on side $AC$ closer to $A$ that divides $AC$ into a ratio of $1:2$. We know that $\angle ABP = 15^{\circ}$. What is the measure of $\angle ACB$?
I. Solution. Drop a perpendicular from point $C$ to segment $PB$, and let the foot of this perpendicular be $E$. In the right triangle $CEP$, $\angle CPE = \angle PAB + \angle PBA = 60^\circ$,

thus $\angle PCE = 30^\circ$, and $PE = \frac{PC}{2}$. Since $PA$ is also equal to $\frac{PC}{2}$, it follows that $PE = PA$, so triangle $APE$ is isosceles. In this triangle,

$$
\angle PAE = \angle PEA = \frac{\angle CPE}{2} = 30^\circ.
$$

Therefore, $CE = EA$. Since $\angle EAB = \angle PAB - \angle PAE = 45^\circ - 30^\circ = 15^\circ$, and $\angle EBA = 15^\circ$, triangle $AEB$ is also isosceles, so $EA = EB$. We have thus found that $CE = EA = EB$, meaning triangle $CEB$ is also isosceles, and since there is a right angle at vertex $E$, $\angle CBE = \angle BCE = 45^\circ$. Based on this, $\angle ACB = \angle PCE + \angle BCE = 30^\circ + 45^\circ = 75^\circ$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_0e0788b0e3b74d5445bbg-1.jpg?height=460&width=553&top_left_y=632&top_left_x=775)

II. Solution. Let $x$ be one-third of side $b$, $T$ be the foot of the perpendicular from point $C$ to side $AB$, and $\delta_1 := \angle BCT$ (Figure 2). Then $\angle TCA$ is also $45^\circ$, and triangle $ATC$ is an isosceles right triangle, so $AT = TC = 3x \cdot \frac{\sqrt{2}}{2}$. Applying the Law of Sines in triangle $ABP$:

$$
\begin{aligned}
\frac{c}{x} & = \frac{\sin 120^\circ}{\sin (45^\circ - 30^\circ)} = \frac{\sin 120^\circ}{\sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ} = \\
& = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}} = \frac{2 \sqrt{3}}{\sqrt{6} - \sqrt{2}} = \frac{2 \sqrt{3} (\sqrt{6} + \sqrt{2})}{4} = \frac{\sqrt{18} + \sqrt{6}}{2} = \frac{3 \sqrt{2} + \sqrt{6}}{2}
\end{aligned}
$$

From this, $c = \frac{3 \sqrt{2} + \sqrt{6}}{2} x$, and $TB = AB - AT = \frac{3 \sqrt{2} + \sqrt{6}}{2} x - 3x \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{2} x$. Writing the tangent of $\delta_1$ in triangle $CTB$:

$$
\tan \delta_1 = \frac{TB}{TC} = \frac{x \cdot \frac{\sqrt{6}}{2}}{3x \cdot \frac{\sqrt{2}}{2}} = \frac{\sqrt{3}}{3}
$$

From this, $\delta_1 = 30^\circ$, so $\angle ABC = 45^\circ + 30^\circ = 75^\circ$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0e0788b0e3b74d5445bbg-1.jpg?height=461&width=553&top_left_y=1938&top_left_x=775)