A geometry problem by Leonardo Joau

Here, $BS=1; SC=\sqrt{2}$ . Now, applying sine rule in triangles ABS and ACS yields $\frac{BS}{sin 30^\circ}=\frac{SC}{sin\angle CAS}$ So, $\angle CAS=45^\circ$ . Triangle CPB is inscribed in a semicircle. Hence, $\angle CPB=\angle APB=90^\circ$ . Finally, from triangle APB, we have $\angle ABP=15^\circ$ .

Let angle $BAC = \theta$ . Let $O$ be the center of $\alpha$ . Draw radius $PO$ of circle $\alpha$ . Since $POC$ is isosceles and with vertex at $O$ , and $ABC$ and $POC$ share angle $ACB$ , we can conclude that $ABC \cong POC$ , therefore $POC = \theta$ . Since angle $PBC$ is the inscribed angle of $POC$ , then we can conclude that $PBC = \frac{POC}{2} = \frac{\theta}{2}$ . But, since angle $ABC = 90-\frac{\theta}{2}$ and $ABP+PBC = ABC$ , then we can equate the two to get $90-\frac{\theta}{2} = ABP+\frac{\theta}{2}$ , giving us $ABP = 90-\theta$ .

Let angle $SAC = \gamma$ . We get that $30+\gamma = \theta$ . By sin law on triangle $ABS$ , $\frac{SB}{\sin 30} = \frac {AB}{\sin ASB}$ . Also, by sin law on triangle $ASC$ , $\frac{SC}{\sin \gamma} = \frac {AC}{\sin ASC}$ . But, since $SB = 1$ , $SC = \sqrt{2}$ , $\sin ASB = \sin ASC$ (sines of supplementary angles are congruent), $AB = AC$ , we can combine the two equations above to give us $\frac{SB}{\sin 30} = \frac{SC}{\sin \gamma}$ , which simplifies to $\sin \gamma = \frac{1}{\sqrt{2}}$ , or $\gamma = 45$ .

Since $\gamma = 45$ , $theta = 30 + \gamma = 30+45 = 75$ . Also, since $ABP = 90 - \theta$ , $ABP = 90 - 75 = 15$ .

Therefore, our answer is $\boxed{15}$ ..

We have $BS=1$ and $SC=\sqrt2$ and $\sin \angle ABC = \sin \angle ACB$ .

By Applying Sine Law on Triangles $ABS$ and $ACS$ ,

$\dfrac{\sin 30^\circ}{1}=\dfrac{\sin \angle ABS}{AS}=\dfrac{\sin \angle ACS}{AS}=\dfrac{\sin \angle SAC}{\sqrt2}$

Therefore $\angle SAC=45^\circ$ . Let $M$ be the point of intersection of $AS$ and $BC$ . We see that $\angle AMP$ and $\angle AMB$ are supplementary. Since $P$ is a point on $\alpha$ , $\angle APB=\angle BPC=90^\circ$ . This gives us $\angle AMP = 45^\circ$ , $\angle AMB=135^\circ$ and $\angle ABP=\boxed{15^\circ}$